Maximize value of a pair from two given arrays based on given conditions
Given two arrays A[] and B[] consisting of N integers and an integer K, the task is to find the maximum value of B[i] + B[j] + abs(A[i] – A[j]) by choosing any pair (i, j) such that abs(A[i] – A[j]) ≤ K.
Examples:
Input: A[] = {5, 6, 9, 10}, B[] = {3, 0, 10, -10}, K = 1
Output: 4
Explanation:
Only two pairs can be chosen, i.e. (0, 1) and (2, 3), because abs(A[0] – A[1]) ≤ K and abs(A[2] – A[3]) ≤ K.
The value of (0, 1) pair is B[0] + B[1] + abs(A[0] – A[1]) = 3 + 0 + 1 = 4.
The value of (2, 3) pair is B[2] + B[3] + abs(A[2] – A[3]) = 10 + (-10) + 1 = 1.
Hence, the maximum value from all possible pairs is 4.Input: A[] = {1, 2, 3, 4}, B[] = {0, 8, 6, 9}, K = 2
Output: 19
Naive Approach: The simplest approach is to generate all possible pairs from the given array and count those pairs that satisfy the given conditions. After checking all the pairs print the count of all pairs.
Time Complexity: O(N2)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to use Segment Trees, Binary Search, and Sorting of the array according to the value of array A[]. Observe that, from the given equation it is clear that B[i] + B[j] + abs(A[i] – A[j]) equals to any of the below values:
- B[i] + B[j] + (A[i] – A[j])
- B[i] + B[j] + (A[j] – A[i])
Consider 2nd equation:
B[i] + B[j] + A[j] – A[i] = B[i] – A[i] + (B[j] + A[j])
- Here, it is observed that for each i in the array A[], finding the right most index of value smaller than of equal to A[i] + K. Let the rightmost index be right.
- Now, calculate B[i] – A[i] + max value of B[j] + A[j] where i + 1 <= j <= right. This will give the maximum value of selected pair.
- To calculate the maximum value each time in a range, segment trees can be used.
Follow the below steps to solve the problem:
- Sort all values, B[i] and B[i] + A[i] according to the values of array A[].
- Initialize maxValue as INT_MIN to store the final maximum answer and initialize a range maximum query segment tree storing all values, A[i] + B[i].
- Traverse the array A[] and perform the following:
- For each element find the index, say right, such that abs(A[i] – A[right]) <= K using Binary Search.
- Then, find the maximum value, of (A[j] + B[j]) where i+1 <= j <= right.
- Update maxValue as maxValue = max(maxValue, B[i] – A[i] + A[j] + B[j]).
- After the above steps, print the value of maxValue as the result.
Below is the implementation of the above approach:
Java
// Java program for the above approachimport java.util.*;// Class to store the values of a[i],// b[i], a[i] + b[i]class triplet implements Comparable<triplet> { int a, b, c; // Constructor triplet(int a, int b, int c) { this.a = a; this.b = b; this.c = c; } // Sort values according to array public int compareTo(triplet o) { return this.a - o.a; }}class GFG { // Stores the segment tree static int seg[]; // Function to find the maximum // possible value according to the // given condition public static void maxValue( int a[], int b[], int n, int k) { // Initialize triplet array triplet arr[] = new triplet[n]; // Store the values a[i], b[i], // a[i] + b[i] for (int i = 0; i < n; i++) { arr[i] = new triplet(a[i], b[i], a[i] + b[i]); } // Sort the array according // to array a[] Arrays.sort(arr); // Build segment tree seg = new int[4 * n]; build(arr, 0, 0, n - 1); // Initialise the maxvalue of // the selected pairs int maxvalue = Integer.MIN_VALUE; // Traverse the array for (int i = 0; i < n; i++) { // For each value find the // floor(arr[i] + k) int right = search(arr, arr[i].a + k); // Find the maximum value of // the select pairs i and max // from (i + 1, right) if (right != -1) { maxvalue = Math.max( maxvalue, arr[i].b - arr[i].a + getMax(arr, 0, 0, n - 1, i + 1, right)); } } // Print the maximum value System.out.println(maxvalue); } // Function to search floor of // the current value public static int search( triplet arr[], int val) { // Initialise low and high values int low = 0, high = arr.length - 1; int ans = -1; // Perform Binary Search while (low <= high) { int mid = low + (high - low) / 2; // If the current value is // <= val then store the // candidate answer and // find for rightmost one if (arr[mid].a <= val) { ans = mid; low = mid + 1; } else high = mid - 1; } return ans; } // Function to build segment tree public static void build( triplet arr[], int index, int s, int e) { // Base Case if (s == e) { seg[index] = arr[s].c; return; } int mid = s + (e - s) / 2; // Build the left and right // segment trees build(arr, 2 * index + 1, s, mid); build(arr, 2 * index + 2, mid + 1, e); // Update current index seg[index] = Math.max(seg[2 * index + 1], seg[2 * index + 2]); } // Function to get maximum value // in the range [qs, qe] public static int getMax( triplet arr[], int index, int s, int e, int qs, int qe) { // If segement is not in range if (qe < s || e < qs) return Integer.MIN_VALUE / 2; // If segement is completely // inside the query if (s >= qs && e <= qe) return seg[index]; // Calculate the maximum value // in left and right half int mid = s + (e - s) / 2; return Math.max( getMax(arr, 2 * index + 1, s, mid, qs, qe), getMax(arr, 2 * index + 2, mid + 1, e, qs, qe)); } // Driver Code public static void main(String args[]) { int N = 4, K = 1; int A[] = { 5, 6, 9, 10 }; int B[] = { 3, 0, 10, -10 }; // Function call maxValue(A, B, N, K); }} |
C#
// C# program for the above approachusing System;using System.Collections.Generic;// Class to store the values of a[i],// b[i], a[i] + b[i]public class triplet : IComparable<triplet> { public int a, b, c; // Constructor public triplet(int a, int b, int c) { this.a = a; this.b = b; this.c = c; } // Sort values according to array public int CompareTo(triplet o) { return this.a - o.a; }}public class GFG { // Stores the segment tree static int []seg; // Function to find the maximum // possible value according to the // given condition public void maxValue( int []a, int []b, int n, int k) { // Initialize triplet array triplet []arr = new triplet[n]; // Store the values a[i], b[i], // a[i] + b[i] for (int i = 0; i < n; i++) { arr[i] = new triplet(a[i], b[i], a[i] + b[i]); } // Sort the array according // to array []a Array.Sort(arr); // Build segment tree seg = new int[4 * n]; build(arr, 0, 0, n - 1); // Initialise the maxvalue of // the selected pairs int maxvalue = int.MinValue; // Traverse the array for (int i = 0; i < n; i++) { // For each value find the // floor(arr[i] + k) int right = search(arr, arr[i].a + k); // Find the maximum value of // the select pairs i and max // from (i + 1, right) if (right != -1) { maxvalue = Math.Max( maxvalue, arr[i].b - arr[i].a + getMax(arr, 0, 0, n - 1, i + 1, right)); } } // Print the maximum value Console.WriteLine(maxvalue); } // Function to search floor of // the current value public int search( triplet []arr, int val) { // Initialise low and high values int low = 0, high = arr.Length - 1; int ans = -1; // Perform Binary Search while (low <= high) { int mid = low + (high - low) / 2; // If the current value is // <= val then store the // candidate answer and // find for rightmost one if (arr[mid].a <= val) { ans = mid; low = mid + 1; } else high = mid - 1; } return ans; } // Function to build segment tree public static void build( triplet []arr, int index, int s, int e) { // Base Case if (s == e) { seg[index] = arr[s].c; return; } int mid = s + (e - s) / 2; // Build the left and right // segment trees build(arr, 2 * index + 1, s, mid); build(arr, 2 * index + 2, mid + 1, e); // Update current index seg[index] = Math.Max(seg[2 * index + 1], seg[2 * index + 2]); } // Function to get maximum value // in the range [qs, qe] public static int getMax( triplet []arr, int index, int s, int e, int qs, int qe) { // If segement is not in range if (qe < s || e < qs) return int.MinValue / 2; // If segement is completely // inside the query if (s >= qs && e <= qe) return seg[index]; // Calculate the maximum value // in left and right half int mid = s + (e - s) / 2; return Math.Max( getMax(arr, 2 * index + 1, s, mid, qs, qe), getMax(arr, 2 * index + 2, mid + 1, e, qs, qe)); } // Driver Code public static void Main(String []args) { int N = 4, K = 1; int []A = { 5, 6, 9, 10 }; int []B = { 3, 0, 10, -10 }; // Function call new GFG().maxValue(A, B, N, K); }}// This code is contributed by 29AjayKumar |
Time Complexity: O(N*log N)
Auxiliary Space: O(N)
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