Longest subarray such that the difference of max and min is at-most one

• Difficulty Level : Hard
• Last Updated : 26 Aug, 2021

Given an array of n numbers where the difference between each number and the previous one doesn’t exceed one. Find the longest contiguous subarray such that the difference between the maximum and minimum number in the range doesn’t exceed one.

Examples:

Input : {3, 3, 4, 4, 5, 6}
Output : 4
The longest subarray here is {3, 3, 4, 4}

Input : {7, 7, 7}
Output : 3
The longest subarray here is {7, 7, 7}

Input : {9, 8, 8, 9, 9, 10}
Output : 5
The longest subarray here is {9, 8, 8, 9, 9}

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

If the difference between the maximum and the minimum numbers in the sequence doesn’t exceed one then the sequence consisted of only one or two consecutive numbers. The idea is to traverse the array and keep track of current element and previous element in current subarray. If we find an element which is not same as current or previous, we update current and previous. We also update result if required.

C++

 // C++ code to find longest // subarray with difference // between max and min as at-most 1. #include using namespace std;    int longestSubarray(int input[],                      int length) {     int prev = -1;     int current, next;     int prevCount = 0, currentCount = 1;        // longest constant range length     int longest = 1;        // first number     current = input;        for (int i = 1; i < length; i++)     {         next = input[i];            // If we see same number         if (next == current)          {             currentCount++;         }            // If we see different number,          // but same as previous.         else if (next == prev)         {             prevCount += currentCount;             prev = current;             current = next;             currentCount = 1;         }            // If number is neither same          // as previous nor as current.          else          {             longest = max(longest,                            currentCount + prevCount);             prev = current;             prevCount = currentCount;             current = next;             currentCount = 1;         }     }        return max(longest,                 currentCount + prevCount); }    // Driver Code int main() {     int input[] = { 5, 5, 6, 7, 6 };      int n = sizeof(input) / sizeof(int);     cout << longestSubarray(input, n);     return 0; }    // This code is contributed // by Arnab Kundu

Java

 // Java code to find longest subarray with difference // between max and min as at-most 1. public class Demo {        public static int longestSubarray(int[] input)     {         int prev = -1;         int current, next;         int prevCount = 0, currentCount = 1;            // longest constant range length         int longest = 1;            // first number         current = input;            for (int i = 1; i < input.length; i++) {             next = input[i];                // If we see same number             if (next == current) {                 currentCount++;             }                // If we see different number, but             // same as previous.             else if (next == prev) {                 prevCount += currentCount;                 prev = current;                 current = next;                 currentCount = 1;             }                // If number is neither same as previous             // nor as current.              else {                 longest = Math.max(longest, currentCount + prevCount);                 prev = current;                 prevCount = currentCount;                 current = next;                 currentCount = 1;             }         }            return Math.max(longest, currentCount + prevCount);     }        public static void main(String[] args)     {         int[] input = { 5, 5, 6, 7, 6 };         System.out.println(longestSubarray(input));     } }

Python 3

 # Python 3 code to find longest # subarray with difference # between max and min as at-most 1. def longestSubarray(input, length):        prev = -1     prevCount = 0     currentCount = 1        # longest constant range length     longest = 1        # first number     current = input        for i in range(1, length):            next = input[i]            # If we see same number         if next == current :             currentCount += 1                # If we see different number,          # but same as previous.         elif next == prev :             prevCount += currentCount             prev = current             current = next             currentCount = 1                    # If number is neither same          # as previous nor as current.          else:             longest = max(longest,                            currentCount +                            prevCount)             prev = current             prevCount = currentCount             current = next             currentCount = 1                return max(longest,              currentCount + prevCount)    # Driver Code if __name__ == "__main__":     input = [ 5, 5, 6, 7, 6 ]     n = len(input)     print(longestSubarray(input, n))        # This code is contributed # by ChitraNayal

C#

 // C# code to find longest  // subarray with difference  // between max and min as  // at-most 1. using System;    class GFG { public static int longestSubarray(int[] input) {     int prev = -1;     int current, next;     int prevCount = 0,          currentCount = 1;        // longest constant      // range length     int longest = 1;        // first number     current = input;        for (int i = 1;               i < input.Length; i++)     {         next = input[i];            // If we see same number         if (next == current)          {             currentCount++;         }            // If we see different number,          // but same as previous.         else if (next == prev)          {             prevCount += currentCount;             prev = current;             current = next;             currentCount = 1;         }            // If number is neither          // same as previous         // nor as current.          else          {             longest = Math.Max(longest,                                 currentCount +                                 prevCount);             prev = current;             prevCount = currentCount;             current = next;             currentCount = 1;         }     }        return Math.Max(longest,                      currentCount + prevCount); }    // Driver Code public static void Main(String[] args) {     int[] input = {5, 5, 6, 7, 6};     Console.WriteLine(longestSubarray(input)); } }    // This code is contributed // by Kirti_Mangal



Javascript



Output:

3

Time Complexity: O(n) where n is the length of the input array.

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