Maximum modulo of all the pairs of array where arr[i]>=arr[j]

Given an array arr[] of n integers. Find the maximum value of arr[i] mod arr[j] where arr[i] >= arr[j] and 1 <= i, j <= n

Examples:

Input: arr[] = {3, 4, 7}
Output: 3
Explanation: There are 3 pairs which satisfiy arr[i] >= arr[j] are:- {4, 3} , {7, 3} and {7, 4}. For which the Maximum modulo value among all is 3 ( 7%4 =3).

Input: arr[] = {3, 7, 4, 11}
Output: 4
Explanation: There are 6 pairs which satisfiy arr[i] >= arr[j] are:- {4, 3} , {7, 3}, {11, 3}, {7, 4}, {11, 4}, {11, 7} . For which the Maximum modulo value among all is 4 (11%7 = 4).

Input: arr[] = {4, 4, 4}
Output: 0
Explanation: Since all the values of the array are equal, the Maximum modulo value is 0.

Try It Yourself

Table of Content

[Naive Approach] - Using Nested Loops - O(n2) Time and O(1) Space
[Expected Approach] - Using Sorting and Binary Search - O(nlog(n) + Mlog(M)) Time and O(1) Space

[Naive Approach] - Using Nested Loops - O(n²) Time and O(1) Space

The idea for this approach is to run two nested for loops and select the maximum of every possible pairs after taking modulo of them. Time complexity of this approach will be O(n²) which will not be sufficient for large value of n.

C++

#include <bits/stdc++.h>
using namespace std;

int maxModValue(vector<int>& arr) {
    int n = arr.size();
    int ans = 0;
    
    // Two nested loops to check every pair (i, j)
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (arr[i] >= arr[j]) {  // Only calculate mod if arr[i] >= arr[j]
                int currMod = arr[i] % arr[j];
                ans = max(ans, currMod);
            }
        }
    }
    
    return ans;
}

int main() {
    vector<int> arr = { 3, 4, 5, 9, 11 };
    cout << maxModValue(arr) << endl;
    return 0;
}

Java

public class GfG {
    public static int maxModValue(int[] arr) {
        int n = arr.length;
        int ans = 0;
        
        // Two nested loops to check every pair (i, j)
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (arr[i] >= arr[j]) {  // Only calculate mod if arr[i] >= arr[j]
                    int currMod = arr[i] % arr[j];
                    ans = Math.max(ans, currMod);
                }
            }
        }
        
        return ans;
    }

    public static void main(String[] args) {
        int[] arr = { 3, 4, 5, 9, 11 };
        System.out.println(maxModValue(arr));
    }
}

Python

def maxModValue(arr):
    n = len(arr)
    ans = 0
    
    # Two nested loops to check every pair (i, j)
    for i in range(n):
        for j in range(n):
            if arr[i] >= arr[j]:  # Only calculate mod if arr[i] >= arr[j]
                currMod = arr[i] % arr[j]
                ans = max(ans, currMod)
    
    return ans

if __name__ == '__main__':
    arr = [3, 4, 5, 9, 11]
    print(maxModValue(arr))

using System;
using System.Linq;

public class GfG {
    public static int maxModValue(int[] arr) {
        int n = arr.Length;
        int ans = 0;
        
        // Two nested loops to check every pair (i, j)
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (arr[i] >= arr[j]) {  // Only calculate mod if arr[i] >= arr[j]
                    int currMod = arr[i] % arr[j];
                    ans = Math.Max(ans, currMod);
                }
            }
        }
        
        return ans;
    }

    public static void Main() {
        int[] arr = { 3, 4, 5, 9, 11 };
        Console.WriteLine(maxModValue(arr));
    }
}

JavaScript

function maxModValue(arr) {
    let n = arr.length;
    let ans = 0;
    
    // Two nested loops to check every pair (i, j)
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {
            if (arr[i] >= arr[j]) {  // Only calculate mod if arr[i] >= arr[j]
                let currMod = arr[i] % arr[j];
                ans = Math.max(ans, currMod);
            }
        }
    }
    
    return ans;
}

const arr = [3, 4, 5, 9, 11];
console.log(maxModValue(arr));

Output

[Expected Approach for Small Range] - Using Sorting and Binary Search - O(nlog(n) + Mlog(M)) Time and O(1) Space

The idea for this approach is to sort the array and then use binary search. For each arr[j], iterate through multiples of arr[j] in the range from 2 * arr[j] to M + arr[j] (where M is the maximum value in the sequence). For each multiple x, use binary search to find the largest arr[i] such that arr[i] < x. This ensures we choose values of arr[i] that maximize arr[i] % arr[j]. Repeat the process for each arr[j] and update the result accordingly.

Example:

If arr[] = {4, 6, 7, 8, 10, 12, 15} then for
first element, i.e., arr[j] = 4 we iterate
through x = {8, 12, 16}.
Therefore for each value of x, a[i] will be:-
x = 8, arr[i] = 7 (7 < 8)
ans = 7 mod 4 = 3
x = 12, arr[i] = 10 (10 < 12)
ans = 10 mod 4 = 2 (Since 2 < 3, No update)
x = 16, arr[i] = 15 (15 < 16)
ans = 15 mod 4 = 3 (Since 3 == 3, No need to update)

C++

#include <bits/stdc++.h>
using namespace std;

int maxModValue(vector<int>& arr)
{
    int ans = 0;
    int n = arr.size();
    
    // Sort the vector by using inbuilt sort function
    sort(arr.begin(), arr.end());

    for (int j = n - 2; j >= 0; --j) {
        
        // Break loop if answer is greater or equals to
        // the arr[j] as any number modulo with arr[j]
        // can only give maximum value up-to arr[j]-1
        if (ans >= arr[j])
            break;

        // If both elements are same then skip the next
        // loop as it would be worthless to repeat the
        // rest process for same value
        if (arr[j] == arr[j + 1])
            continue;

        for (int i = 2 * arr[j]; i <= arr[n - 1] + arr[j]; i += arr[j]) {
            
            // Fetch the index which is greater than or
            // equals to arr[i] by using binary search
            // inbuilt lower_bound() function of C++
            int ind = lower_bound(arr.begin(), arr.end(), i) - arr.begin();

            // Update the answer
            ans = max(ans, arr[ind - 1] % arr[j]);
        }
    }
    return ans;
}

// Driver code
int main()
{
    vector<int> arr = { 3, 4, 5, 9, 11 };
    cout << maxModValue(arr);
}