Given an array arr[] of N positive integers. The task is to find the maximum possible value of a[i] % a[j] over all pairs of i and j.
Examples:
Input: arr[] = {4, 5, 1, 8}
Output: 5
If we choose the pair (5, 8), then 5 % 8 gives us 5
which is the maximum possible.
Input: arr[] = {7, 7, 8, 8, 1}
Output: 7
Approach: Since we can choose any pair, hence arr[i] should be the second maximum of the array and arr[j] be the maximum element in order to maximize the required value. Hence, the second maximum over the array will be our answer. If there does not exist any second largest number, then 0 will be the answer.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function that returns the second largest// element in the array if exists, else 0int getMaxValue(int arr[], int arr_size){ int i, first, second; // There must be at least two elements if (arr_size < 2) { return 0; } // To store the maximum and the second // maximum element from the array first = second = INT_MIN; for (i = 0; i < arr_size; i++) { // If current element is greater than first // then update both first and second if (arr[i] > first) { second = first; first = arr[i]; } // If arr[i] is in between first and // second then update second else if (arr[i] > second && arr[i] != first) second = arr[i]; } // No second maximum found if (second == INT_MIN) return 0; else return second;}// Driver codeint main(){ int arr[] = { 4, 5, 1, 8 }; int n = sizeof(arr) / sizeof(arr[0]); cout << getMaxValue(arr, n); return 0;} |
Java
// Java implementation of the approachclass GFG{ // Function that returns the second largest // element in the array if exists, else 0 static int getMaxValue(int arr[], int arr_size) { int i, first, second; // There must be at least two elements if (arr_size < 2) { return 0; } // To store the maximum and the second // maximum element from the array first = second = Integer.MIN_VALUE; for (i = 0; i < arr_size; i++) { // If current element is greater than first // then update both first and second if (arr[i] > first) { second = first; first = arr[i]; } // If arr[i] is in between first and // second then update second else if (arr[i] > second && arr[i] != first) { second = arr[i]; } } // No second maximum found if (second == Integer.MIN_VALUE) { return 0; } else { return second; } } // Driver code public static void main(String[] args) { int arr[] = {4, 5, 1, 8}; int n = arr.length; System.out.println(getMaxValue(arr, n)); }} // This code has been contributed by 29AjayKumar |
Python3
import sys# Python 3 implementation of the approach# Function that returns the second largest# element in the array if exists, else 0def getMaxValue(arr,arr_size): # There must be at least two elements if (arr_size < 2): return 0 # To store the maximum and the second # maximum element from the array first = -sys.maxsize-1 second = -sys.maxsize-1 for i in range(arr_size): # If current element is greater than first # then update both first and second if (arr[i] > first): second = first first = arr[i] # If arr[i] is in between first and # second then update second elif (arr[i] > second and arr[i] != first): second = arr[i] # No second maximum found if (second == -sys.maxsize-1): return 0 else: return second# Driver codeif __name__ == '__main__': arr = [4, 5, 1, 8] n = len(arr) print(getMaxValue(arr, n))# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approachusing System;class GFG{ // Function that returns the second largest // element in the array if exists, else 0 static int getMaxValue(int []arr, int arr_size) { int i, first, second; // There must be at least two elements if (arr_size < 2) { return 0; } // To store the maximum and the second // maximum element from the array first = second = int.MinValue; for (i = 0; i < arr_size; i++) { // If current element is greater than first // then update both first and second if (arr[i] > first) { second = first; first = arr[i]; } // If arr[i] is in between first and // second then update second else if (arr[i] > second && arr[i] != first) { second = arr[i]; } } // No second maximum found if (second == int.MinValue) { return 0; } else { return second; } } // Driver code public static void Main() { int []arr = {4, 5, 1, 8}; int n = arr.Length; Console.Write(getMaxValue(arr, n)); }} // This code is contributed by Akanksha Rai |
PHP
<?php// PHP implementation of the approach // Function that returns the second largest // element in the array if exists, else 0 function getMaxValue($arr, $arr_size) { // There must be at least two elements if ($arr_size < 2) { return 0; } // To store the maximum and the second // maximum element from the array $first = $second = -(PHP_INT_MAX - 1); for ($i = 0; $i < $arr_size; $i++) { // If current element is greater than first // then update both first and second if ($arr[$i] > $first) { $second = $first; $first = $arr[$i]; } // If arr[i] is in between first and // second then update second else if ($arr[$i] > $second && $arr[$i] != $first) $second = $arr[$i]; } // No second maximum found if ($second == -(PHP_INT_MAX-1)) return 0; else return $second; } // Driver code $arr = array(4, 5, 1, 8); $n = count($arr);echo getMaxValue($arr, $n); // This code is contributed by Ryuga?> |
Javascript
<script>// JavaScript implementation of the approach // Function that returns the second largest// element in the array if exists, else 0 function getMaxValue(arr, arr_size) { let i, first, second; // There must be at least two elements if (arr_size < 2) { return 0; } // To store the maximum and the second // maximum element from the array first = second = Number.MIN_VALUE; for (i = 0; i < arr_size; i++) { // If current element is greater than first // then update both first and second if (arr[i] > first) { second = first; first = arr[i]; } // If arr[i] is in between first and // second then update second else if (arr[i] > second && arr[i] != first) { second = arr[i]; } } // No second maximum found if (second == Number.MIN_VALUE) { return 0; } else { return second; } } // Driver Code let arr = [4, 5, 1, 8]; let n = arr.length; document.write(getMaxValue(arr, n)); </script> |
Output:
5
Time Complexity: O(n) where n is the size of the array
Auxiliary Space: O(1)
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