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# Maximum length of same indexed subarrays from two given arrays satisfying the given condition

• Difficulty Level : Expert
• Last Updated : 11 Jul, 2022

Given two arrays arr[] and brr[] and an integer C, the task is to find the maximum possible length, say K, of the same indexed subarrays such that the sum of the maximum element in the K-length subarray in brr[] with the product between K and sum of the K-length subarray in arr[] does not exceed C.

Examples:

Input: arr[] = {2, 1, 3, 4, 5}, brr[] = {3, 6, 1, 3, 4}, C = 25
Output: 3
Explanation: Considering the subarrays arr[] = {2, 1, 3} (Sum = 6) and brr[] = {3, 6, 1} (Maximum element = 6), Maximum element + sum * K = 6 + 6 * 3 = 24, which is less than C(= 25).

Input: arr[] ={1, 2, 1, 6, 5, 5, 6, 1}, brr[] = {14, 8, 15, 15, 9, 10, 7, 12}, C = 40
Output: 3
Explanation: Considering the subarrays arr[] = {1, 2, 1} (Sum = 4) and brr[] = {14, 8, 15} (Maximum element = 6), Maximum element + sum * K = 15 + 4 * 3 = 27, which is less than C(= 40).

Naive Approach: The simplest approach is to generate all possible subarrays of the two given arrays and consider all similarly indexed subarrays from both the arrays and check for the given condition. Print the maximum length of subarrays satisfying the given conditions.

Time Complexity: O(K*N2)
Auxiliary Space: O(1)

Binary-Search-based Approach: To optimize the above approach, the idea is to use Binary Search to find the possible value of K and to find the sum of each subarray of length K using the Sliding Window Technique. Follow the steps below to solve the problem:

• Build a Segment Tree to find the maximum value among all possible ranges.
• Perform Binary Search over the range [0, N] to find the maximum possible size of the subarray.
• Initialize low as 0 and high as N.
• Find the value of mid as (low + high)/2.
• Check if it is possible to get the maximum size of the subarray as mid or not by checking the given condition. If found to be true, then update the maximum length as mid and low as (mid + 1).
• Otherwise, update high as (mid – 1).
• After completing the above steps, print the value of the maximum length stored.

Below is the implementation of the above approach:

## C++14

 // C++ program for the above approach  #include   using namespace std;     // Stores the segment tree node values int seg[10000];    // Function to find maximum element // in the given range int getMax(int b[], int ss, int se, int qs,            int qe, int index) {            // If the query is out of bounds     if (se < qs || ss > qe)         return INT_MIN / 2;        // If the segment is completely     // inside the query range     if (ss >= qs && se <= qe)         return seg[index];        // Calculate the mid     int mid = ss + (se - ss) / 2;        // Return maximum in left & right     // of the segment tree recursively     return max(         getMax(b, ss, mid, qs,                qe, 2 * index + 1),         getMax(b, mid + 1, se,                qs, qe, 2 * index + 2)); }    // Function to check if it is possible // to have such a subarray of length K bool possible(int a[], int b[], int n,               int c, int k) {     int sum = 0;            // Check for first window of size K     for(int i = 0; i < k; i++)      {         sum += a[i];     }         // Calculate the total cost and     // check if less than equal to c     int total_cost = sum * k + getMax(                      b, 0, n - 1,                          0, k - 1, 0);         // If it satisfy the condition     if (total_cost <= c)         return true;         // Find the sum of current subarray     // and calculate total cost     for(int i = k; i < n; i++)     {                    // Include the new element         // of current subarray         sum += a[i];             // Discard the element         // of last subarray         sum -= a[i - k];             // Calculate total cost         // and check <=c         total_cost = sum * k + getMax(                      b, 0, n - 1,                        i - k + 1, i, 0);             // If possible, then         // return true         if (total_cost <= c)             return true;     }         // If it is not possible     return false; }    // Function to find maximum length // of subarray such that sum of // maximum element in subarray in brr[] and // sum of subarray in arr[] * K is at most C int maxLength(int a[], int b[], int n, int c) {            // Base Case     if (n == 0)         return 0;        // Let maximum length be 0     int max_length = 0;         int low = 0, high = n;         // Perform Binary search     while (low <= high)     {                    // Find mid value         int mid = low + (high - low) / 2;                    // Check if the current mid         // satisfy the given condition         if (possible(a, b, n, c, mid) != false)         {                            // If yes, then store length             max_length = mid;             low = mid + 1;         }             // Otherwise         else             high = mid - 1;     }         // Return maximum length stored     return max_length; }     // Function that builds segment Tree void build(int b[], int index, int s, int e) {            // If there is only one element     if (s == e)      {         seg[index] = b[s];         return;     }         // Find the value of mid     int mid = s + (e - s) / 2;        // Build left and right parts     // of segment tree recursively     build(b, 2 * index + 1, s, mid);     build(b, 2 * index + 2, mid + 1, e);        // Update the value at current     // index     seg[index] = max(         seg[2 * index + 1],         seg[2 * index + 2]); }    // Function that initializes the // segment Tree void initialiseSegmentTree(int N) {     int seg[4 * N]; }      // Driver Code  int main()  {      int A[] = { 1, 2, 1, 6, 5, 5, 6, 1 };     int B[] = { 14, 8, 15, 15, 9, 10, 7, 12 };             int C = 40;             int N = sizeof(A) / sizeof(A[0]);             // Initialize and Build the     // Segment Tree     initialiseSegmentTree(N);     build(B, 0, 0, N - 1);             // Function Call     cout << (maxLength(A, B, N, C)); }     // This code is contributed by susmitakundugoaldanga

## Java

 // Java program for the above approach    import java.io.*; import java.util.*; class GFG {        // Stores the segment tree node values     static int seg[];        // Function to find maximum length     // of subarray such that sum of     // maximum element in subarray in brr[] and     // sum of subarray in arr[] * K is at most C     public static int maxLength(         int a[], int b[], int n, int c)     {         // Base Case         if (n == 0)             return 0;            // Let maximum length be 0         int max_length = 0;            int low = 0, high = n;            // Perform Binary search         while (low <= high) {                // Find mid value             int mid = low + (high - low) / 2;                // Check if the current mid             // satisfy the given condition             if (possible(a, b, n, c, mid)) {                    // If yes, then store length                 max_length = mid;                 low = mid + 1;             }                // Otherwise             else                 high = mid - 1;         }            // Return maximum length stored         return max_length;     }        // Function to check if it is possible     // to have such a subarray of length K     public static boolean possible(         int a[], int b[], int n,         int c, int k)     {         int sum = 0;            // Check for first window of size K         for (int i = 0; i < k; i++) {             sum += a[i];         }            // Calculate the total cost and         // check if less than equal to c         int total_cost             = sum * k + getMax(b, 0, n - 1,                                0, k - 1, 0);            // If it satisfy the condition         if (total_cost <= c)             return true;            // Find the sum of current subarray         // and calculate total cost         for (int i = k; i < n; i++) {                // Include the new element             // of current subarray             sum += a[i];                // Discard the element             // of last subarray             sum -= a[i - k];                // Calculate total cost             // and check <=c             total_cost                 = sum * k                   + getMax(b, 0, n - 1,                            i - k + 1, i, 0);                // If possible, then             // return true             if (total_cost <= c)                 return true;         }            // If it is not possible         return false;     }        // Function that builds segment Tree     public static void build(         int b[], int index, int s, int e)     {         // If there is only one element         if (s == e) {             seg[index] = b[s];             return;         }            // Find the value of mid         int mid = s + (e - s) / 2;            // Build left and right parts         // of segment tree recursively         build(b, 2 * index + 1, s, mid);         build(b, 2 * index + 2, mid + 1, e);            // Update the value at current         // index         seg[index] = Math.max(             seg[2 * index + 1],             seg[2 * index + 2]);     }        // Function to find maximum element     // in the given range     public static int getMax(         int b[], int ss, int se, int qs,         int qe, int index)     {         // If the query is out of bounds         if (se < qs || ss > qe)             return Integer.MIN_VALUE / 2;            // If the segment is completely         // inside the query range         if (ss >= qs && se <= qe)             return seg[index];            // Calculate the mid         int mid = ss + (se - ss) / 2;            // Return maximum in left & right         // of the segment tree recursively         return Math.max(             getMax(b, ss, mid, qs,                    qe, 2 * index + 1),             getMax(b, mid + 1, se,                    qs, qe, 2 * index + 2));     }        // Function that initializes the     // segment Tree     public static void     initialiseSegmentTree(int N)     {         seg = new int[4 * N];     }        // Driver Code     public static void main(String[] args)     {         int A[] = { 1, 2, 1, 6, 5, 5, 6, 1 };         int B[] = { 14, 8, 15, 15, 9, 10, 7, 12 };            int C = 40;            int N = A.length;            // Initialize and Build the         // Segment Tree         initialiseSegmentTree(N);         build(B, 0, 0, N - 1);            // Function Call         System.out.println(maxLength(A, B, N, C));     } }

## Python3

 # Python3 program for the above approach  import math     # Stores the segment tree node values seg = [0 for x in range(10000)]  INT_MIN = int(-10000000)    # Function to find maximum element # in the given range def getMax(b, ss, se, qs, qe, index):            # If the query is out of bounds     if (se < qs or ss > qe):         return int(INT_MIN / 2)                # If the segment is completely     # inside the query range     if (ss >= qs and se <= qe):         return seg[index]                # Calculate the mid     mid = int(int(ss) + int((se - ss) / 2))            # Return maximum in left & right     # of the segment tree recursively     return max(getMax(b, ss, mid, qs,                       qe, 2 * index + 1),                getMax(b, mid + 1, se, qs,                       qe, 2 * index + 2))    # Function to check if it is possible # to have such a subarray of length K def possible(a,  b, n, c, k):            sum = int(0)            # Check for first window of size K     for i in range(0, k):         sum += a[i]                # Calculate the total cost and     # check if less than equal to c     total_cost = int(sum * k +                getMax(b, 0, n - 1,                         0, k - 1, 0))         # If it satisfy the condition     if (total_cost <= c):         return 1        # Find the sum of current subarray     # and calculate total cost     for i in range (k, n):                    # Include the new element         # of current subarray         sum += a[i]                    # Discard the element         # of last subarray         sum -= a[i - k]            # Calculate total cost         # and check <=c         total_cost = int(sum * k + getMax(                b, 0, n - 1,i - k + 1, i, 0))             # If possible, then         # return true         if (total_cost <= c):             return 1                    # If it is not possible     return 0    # Function to find maximum length # of subarray such that sum of # maximum element in subarray in brr[] and # sum of subarray in arr[] * K is at most C def maxLength(a, b, n, c):            # Base Case     if (n == 0):         return 0        # Let maximum length be 0     max_length = int(0)         low = 0     high = n            # Perform Binary search     while (low <= high):                    # Find mid value         mid = int(low + int((high - low) / 2))                    # Check if the current mid         # satisfy the given condition         if (possible(a, b, n, c, mid) != 0):                            # If yes, then store length             max_length = mid             low = mid + 1                        # Otherwise         else:             high = mid - 1         # Return maximum length stored     return max_length     # Function that builds segment Tree def build(b, index, s, e):            # If there is only one element     if (s == e):         seg[index] = b[s]         return            # Find the value of mid     mid = int(s + int((e - s) / 2))        # Build left and right parts     # of segment tree recursively     build(b, 2 * index + 1, s, mid)     build(b, 2 * index + 2, mid + 1, e)        # Update the value at current     # index     seg[index] = max(seg[2 * index + 1],                      seg[2 * index + 2])    #  Driver Code  A = [ 1, 2, 1, 6, 5, 5, 6, 1 ] B = [ 14, 8, 15, 15, 9, 10, 7, 12 ]         C = int(40) N = len(A)      # Initialize and Build the # Segment Tree build(B, 0, 0, N - 1)    # Function Call print((maxLength(A, B, N, C)))    # This code is contributed by Stream_Cipher

## C#

 // C# program for the above approach  using System;    class GFG{        // Stores the segment tree node values  static int[] seg;     // Function to find maximum length  // of subarray such that sum of  // maximum element in subarray in brr[] and  // sum of subarray in arr[] * K is at most C  static int maxLength(int[] a, int[] b,                      int n, int c)  {             // Base Case      if (n == 0)          return 0;         // Let maximum length be 0      int max_length = 0;         int low = 0, high = n;         // Perform Binary search      while (low <= high)     {                     // Find mid value          int mid = low + (high - low) / 2;             // Check if the current mid          // satisfy the given condition          if (possible(a, b, n, c, mid))         {                             // If yes, then store length              max_length = mid;              low = mid + 1;          }             // Otherwise          else             high = mid - 1;      }         // Return maximum length stored      return max_length;  }     // Function to check if it is possible  // to have such a subarray of length K  static bool possible(int[] a, int[] b, int n,                      int c, int k)  {      int sum = 0;         // Check for first window of size K      for(int i = 0; i < k; i++)     {         sum += a[i];      }         // Calculate the total cost and      // check if less than equal to c      int total_cost = sum * k +               getMax(b, 0, n - 1,                          0, k - 1, 0);         // If it satisfy the condition      if (total_cost <= c)          return true;         // Find the sum of current subarray      // and calculate total cost      for(int i = k; i < n; i++)     {                     // Include the new element          // of current subarray          sum += a[i];             // Discard the element          // of last subarray          sum -= a[i - k];             // Calculate total cost          // and check <=c          total_cost = sum * k +               getMax(b, 0, n - 1,                         i - k + 1, i, 0);             // If possible, then          // return true          if (total_cost <= c)              return true;      }         // If it is not possible      return false;  }     // Function that builds segment Tree  static void build(int[] b, int index,                   int s, int e)  {             // If there is only one element      if (s == e)      {          seg[index] = b[s];          return;      }         // Find the value of mid      int mid = s + (e - s) / 2;         // Build left and right parts      // of segment tree recursively      build(b, 2 * index + 1, s, mid);      build(b, 2 * index + 2, mid + 1, e);         // Update the value at current      // index      seg[index] = Math.Max(          seg[2 * index + 1],          seg[2 * index + 2]);  }     // Function to find maximum element  // in the given range  public static int getMax(int[] b, int ss,                           int se, int qs,                           int qe, int index)  {             // If the query is out of bounds      if (se < qs || ss > qe)          return Int32.MinValue / 2;         // If the segment is completely      // inside the query range      if (ss >= qs && se <= qe)          return seg[index];         // Calculate the mid      int mid = ss + (se - ss) / 2;         // Return maximum in left & right      // of the segment tree recursively      return Math.Max(          getMax(b, ss, mid, qs,                 qe, 2 * index + 1),          getMax(b, mid + 1, se,                 qs, qe, 2 * index + 2));  }     // Function that initializes the  // segment Tree  static void initialiseSegmentTree(int N)  {      seg = new int[4 * N];  }     // Driver Code     static void Main() {     int[] A = { 1, 2, 1, 6, 5, 5, 6, 1 };      int[] B = { 14, 8, 15, 15, 9, 10, 7, 12 };         int C = 40;         int N = A.Length;         // Initialize and Build the      // Segment Tree      initialiseSegmentTree(N);      build(B, 0, 0, N - 1);         // Function Call      Console.WriteLine(maxLength(A, B, N, C));  } }    // This code is contributed by divyesh072019

## Javascript



Output:

3

Time Complexity: O(N*(log N)2)
Auxiliary Space: O(N)

Efficient Approach: To optimize the above approach, the idea is to use a Deque by using a monotone queue such that for each subarray of fixed length we can find a maximum in O(1) time. For any subarray in the range [i, i + K – 1] the value expression to be calculated is given by:

Below are the steps:

• Perform the Binary Search over the range [0, N] to find the maximum possible size of the subarray.
• Initialize low as 0 and high as N.
• Find the value of mid as (low + high)/2.
• Check if it is possible to get the maximum size of the subarray as mid or not as:
• Use deque to find the maximum element in each subarray of size K in the array brr[].
• Find the value of the expression and if it at most C then break out of this condition.
• Else check for all possible subarray size mid and if the value of the expression and if it at most C then break out of this condition.
• Return false if none of the above conditions satisfies.
• If the current mid satisfies the given conditions then update maximum length as mid and low as (mid + 1).
• Else Update high as (mid – 1).
• After the above steps, print the value of the maximum length stored.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std;    // Function to check if it is possible // to have such a subarray of length K bool possible(int a[], int b[], int n, int c,                         int k) {        // Finds the maximum element     // in each window of size k     deque dq;        // Check for window of size K     int sum = 0;        // For all possible subarrays of     // length k     for (int i = 0; i < k; i++)      {         sum += a[i];            // Until deque is empty         while (dq.size() > 0 && b[i] > b[dq.back()])             dq.pop_back();         dq.push_back(i);     }        // Calculate the total cost and     // check if less than equal to c     int total_cost = sum * k + b[dq.front()];     if (total_cost <= c)         return true;        // Find sum of current subarray     // and the total cost     for (int i = k; i < n; i++)     {            // Include the new element         // of current subarray         sum += a[i];            // Discard the element         // of last subarray         sum -= a[i - k];            // Remove all the elements         // in the old window         while (dq.size() > 0 && dq.front() <= i - k)             dq.pop_front();         while (dq.size() > 0 && b[i] > b[dq.back()])             dq.pop_back();               dq.push_back(i);            // Calculate total cost         // and check <=c         total_cost = sum * k + b[dq.front()];            // If current subarray         // length satisfies         if (total_cost <= c)             return true;     }        // If it is not possible     return false; }    // Function to find maximum length // of subarray such that sum of // maximum element in subarray in brr[] and // sum of subarray in arr[] * K is at most C int maxLength(int a[], int b[], int n, int c) {          // Base Case     if (n == 0)         return 0;        // Let maximum length be 0     int max_length = 0;     int low = 0, high = n;        // Perform Binary search     while (low <= high)     {            // Find mid value         int mid = low + (high - low) / 2;            // Check if the current mid         // satisfy the given condition         if (possible(a, b, n, c, mid))          {                // If yes, then store length             max_length = mid;             low = mid + 1;         }            // Otherwise         else             high = mid - 1;     }        // Return maximum length stored     return max_length; }    // Driver Code int main() {        int A[] = { 1, 2, 1, 6, 5, 5, 6, 1 };     int B[] = { 14, 8, 15, 15, 9, 10, 7, 12 };     int N = sizeof(A)/sizeof(A[0]);     int C = 40;     cout << maxLength(A, B, N, C);     return 0; }    // This code is contributed by Dharanendra L V

## Java

 // Java program for the above approach    import java.io.*; import java.util.*; class GFG {        // Function to find maximum length     // of subarray such that sum of     // maximum element in subarray in brr[] and     // sum of subarray in arr[] * K is at most C     public static int maxLength(         int a[], int b[], int n, int c)     {         // Base Case         if (n == 0)             return 0;            // Let maximum length be 0         int max_length = 0;            int low = 0, high = n;            // Perform Binary search         while (low <= high) {                // Find mid value             int mid = low + (high - low) / 2;                // Check if the current mid             // satisfy the given condition             if (possible(a, b, n, c, mid)) {                    // If yes, then store length                 max_length = mid;                 low = mid + 1;             }                // Otherwise             else                 high = mid - 1;         }            // Return maximum length stored         return max_length;     }        // Function to check if it is possible     // to have such a subarray of length K     public static boolean possible(         int a[], int b[], int n, int c, int k)     {            // Finds the maximum element         // in each window of size k         Deque dq             = new LinkedList();            // Check for window of size K         int sum = 0;            // For all possible subarrays of         // length k         for (int i = 0; i < k; i++) {                sum += a[i];                // Until deque is empty             while (dq.size() > 0                    && b[i] > b[dq.peekLast()])                 dq.pollLast();             dq.addLast(i);         }            // Calculate the total cost and         // check if less than equal to c         int total_cost = sum * k                          + b[dq.peekFirst()];         if (total_cost <= c)             return true;            // Find sum of current subarray         // and the total cost         for (int i = k; i < n; i++) {                // Include the new element             // of current subarray             sum += a[i];                // Discard the element             // of last subarray             sum -= a[i - k];                // Remove all the elements             // in the old window             while (dq.size() > 0                    && dq.peekFirst()                           <= i - k)                 dq.pollFirst();                while (dq.size() > 0                    && b[i]                           > b[dq.peekLast()])                 dq.pollLast();                dq.add(i);                // Calculate total cost             // and check <=c             total_cost = sum * k                          + b[dq.peekFirst()];                // If current subarray             // length satisfies             if (total_cost <= c)                 return true;         }            // If it is not possible         return false;     }        // Driver Code     public static void main(String[] args)     {         int A[] = { 1, 2, 1, 6, 5, 5, 6, 1 };         int B[] = { 14, 8, 15, 15, 9, 10, 7, 12 };            int N = A.length;            int C = 40;            System.out.println(             maxLength(A, B, N, C));     } }

## Python3

 # Python3 program for the above approach    # Function to find maximum length # of subarray such that sum of # maximum element in subarray in brr[] and # sum of subarray in []arr * K is at most C def maxLength(a, b, n, c):        # Base Case     if(n == 0):         return 0        # Let maximum length be 0     max_length = 0     low = 0     high = n        # Perform Binary search     while(low <= high):            # Find mid value         mid = int(low + (high - low) / 2)            # Check if the current mid         # satisfy the given condition         if(possible(a, b, n, c, mid)):                # If yes, then store length             max_length = mid             low = mid + 1            # Otherwise         else:             high = mid - 1        # Return maximum length stored     return max_length    # Function to check if it is possible # to have such a subarray of length K def possible(a, b, n, c, k):        # Finds the maximum element     # in each window of size k     dq = []            # Check for window of size K     Sum = 0        # For all possible subarrays of     # length k     for i in range(k):         Sum += a[i]            # Until deque is empty         while(len(dq) > 0 and b[i] > b[dq[len(dq) - 1]]):             dq.pop(len(dq) - 1)         dq.append(i)        # Calculate the total cost and     # check if less than equal to c     total_cost = Sum * k + b[dq[0]]     if(total_cost <= c):         return True        # Find sum of current subarray     # and the total cost     for i in range(k, n):            # Include the new element         # of current subarray         Sum += a[i]            # Discard the element         # of last subarray         Sum -= a[i - k]            # Remove all the elements         # in the old window         while(len(dq) > 0 and dq[0] <= i - k):             dq.pop(0)         while(len(dq) > 0 and b[i] > b[dq[len(dq) - 1]]):             dq.pop(len(dq) - 1)         dq.append(i)            # Calculate total cost         # and check <=c         total_cost = Sum * k + b[dq[0]]            # If current subarray         # length satisfies         if(total_cost <= c):             return True            # If it is not possible     return False    # Driver Code A = [1, 2, 1, 6, 5, 5, 6, 1] B = [14, 8, 15, 15, 9, 10, 7, 12] N = len(A) C = 40 print(maxLength(A, B, N, C))    # This code is contributed by avanitrachhadiya2155

## C#

 // C# program for the above approach using System; using System.Collections.Generic;    public class GFG {        // Function to find maximum length     // of subarray such that sum of     // maximum element in subarray in brr[] and     // sum of subarray in []arr * K is at most C     public static int maxLength(         int []a, int []b, int n, int c)     {         // Base Case         if (n == 0)             return 0;            // Let maximum length be 0         int max_length = 0;            int low = 0, high = n;            // Perform Binary search         while (low <= high) {                // Find mid value             int mid = low + (high - low) / 2;                // Check if the current mid             // satisfy the given condition             if (possible(a, b, n, c, mid)) {                    // If yes, then store length                 max_length = mid;                 low = mid + 1;             }                // Otherwise             else                 high = mid - 1;         }            // Return maximum length stored         return max_length;     }        // Function to check if it is possible     // to have such a subarray of length K     public static bool possible(         int []a, int []b, int n, int c, int k)     {            // Finds the maximum element         // in each window of size k         List dq             = new List();            // Check for window of size K         int sum = 0;            // For all possible subarrays of         // length k         for (int i = 0; i < k; i++) {                sum += a[i];                // Until deque is empty             while (dq.Count > 0                    && b[i] > b[dq[dq.Count - 1]])                 dq.RemoveAt(dq.Count - 1);             dq.Add(i);         }            // Calculate the total cost and         // check if less than equal to c         int total_cost = sum * k                          + b[dq[0]];         if (total_cost <= c)             return true;            // Find sum of current subarray         // and the total cost         for (int i = k; i < n; i++) {                // Include the new element             // of current subarray             sum += a[i];                // Discard the element             // of last subarray             sum -= a[i - k];                // Remove all the elements             // in the old window             while (dq.Count > 0                    && dq[0]                           <= i - k)                 dq.RemoveAt(0);                while (dq.Count > 0                    && b[i]                           > b[dq[dq.Count - 1]])                 dq.RemoveAt(dq.Count - 1);                dq.Add(i);                // Calculate total cost             // and check <=c             total_cost = sum * k                          + b[dq[0]];                // If current subarray             // length satisfies             if (total_cost <= c)                 return true;         }            // If it is not possible         return false;     }        // Driver Code     public static void Main(String[] args)     {         int []A = { 1, 2, 1, 6, 5, 5, 6, 1 };         int []B = { 14, 8, 15, 15, 9, 10, 7, 12 };            int N = A.Length;            int C = 40;            Console.WriteLine(             maxLength(A, B, N, C));     } }    // This code is contributed by Amit Katiyar

## Javascript



Output:

3

Time Complexity: O(N*log N)
Auxiliary Space: O(N)

Related Topic: Subarrays, Subsequences, and Subsets in Array

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