Given an array arr[] of N integers, the task is to find the maximum mod value for any pair (arr[i], arr[j]) from the array.
Examples:
Input: arr[] = {2, 4, 1, 5, 3, 6}
Output: 5
(5 % 6) = 5 is the maximum possible mod value.
Input: arr[] = {6, 6, 6, 6}
Output: 0
Approach: It is known that when an integer is divided by some other integer X, the remainder will always be less than X. So, the maximum mod value which can be obtained from the array will be when the divisor is the maximum element from the array and this value will be maximum when the dividend is the maximum among the remaining elements i.e. the second maximum element from the array which is the required answer. Note that the result will be 0 when all the elements of the array are equal.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the maximum mod// value for any pair from the arrayint maxMod(int arr[], int n){ int maxVal = *max_element(arr, arr + n); int secondMax = 0; // Find the second maximum // element from the array for (int i = 0; i < n; i++) { if (arr[i] < maxVal && arr[i] > secondMax) { secondMax = arr[i]; } } return secondMax;}// Driver codeint main(){ int arr[] = { 2, 4, 1, 5, 3, 6 }; int n = sizeof(arr) / sizeof(int); cout << maxMod(arr, n); return 0;} |
Java
// Java implementation of the approach class GFG { static int max_element(int arr[], int n) { int max = arr[0]; for(int i = 1; i < n ; i++) { if (max < arr[i]) max = arr[i]; } return max; } // Function to return the maximum mod // value for any pair from the array static int maxMod(int arr[], int n) { int maxVal = max_element(arr, n); int secondMax = 0; // Find the second maximum // element from the array for (int i = 0; i < n; i++) { if (arr[i] < maxVal && arr[i] > secondMax) { secondMax = arr[i]; } } return secondMax; } // Driver code public static void main (String[] args) { int arr[] = { 2, 4, 1, 5, 3, 6 }; int n = arr.length; System.out.println(maxMod(arr, n)); } }// This code is contributed by AnkitRai01 |
Python3
# Python3 implementation of the approach # Function to return the maximum mod # value for any pair from the array def maxMod(arr, n): maxVal = max(arr) secondMax = 0 # Find the second maximum # element from the array for i in range(0, n): if (arr[i] < maxVal and arr[i] > secondMax): secondMax = arr[i] return secondMax # Driver code arr = [2, 4, 1, 5, 3, 6]n = len(arr)print(maxMod(arr, n))# This code is contributed # by Sanjit Prasad |
C#
// C# implementation of the approachusing System;using System.Collections.Generic; class GFG { static int max_element(int []arr, int n) { int max = arr[0]; for(int i = 1; i < n ; i++) { if (max < arr[i]) max = arr[i]; } return max; } // Function to return the maximum mod // value for any pair from the array static int maxMod(int []arr, int n) { int maxVal = max_element(arr, n); int secondMax = 0; // Find the second maximum // element from the array for (int i = 0; i < n; i++) { if (arr[i] < maxVal && arr[i] > secondMax) { secondMax = arr[i]; } } return secondMax; } // Driver code public static void Main (String[] args) { int []arr = { 2, 4, 1, 5, 3, 6 }; int n = arr.Length; Console.WriteLine(maxMod(arr, n)); } }// This code is contributed by 29AjayKumar |
Javascript
<script>// JavaScript implementation of the approach// Function to return the maximum mod// value for any pair from the arrayfunction maxMod(arr, n) { let maxVal = arr.sort((a, b) => b - a)[0] let secondMax = 0; // Find the second maximum // element from the array for (let i = 0; i < n; i++) { if (arr[i] < maxVal && arr[i] > secondMax) { secondMax = arr[i]; } } return secondMax;}// Driver codelet arr = [2, 4, 1, 5, 3, 6];let n = arr.length;document.write(maxMod(arr, n));</script> |
5
Time Complexity: O(N).
Auxiliary Space: O(1).
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