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

Given an array 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 satisfies arr[i] >= arr[j] are:-
4, 3 => 4 % 3 = 1
7, 3 => 7 % 3 = 1
7, 4 => 7 % 4 = 3
Hence Maximum value among all is 3.

Input: arr[] = {3, 7, 4, 11}
Output: 4

Input: arr[] = {4, 4, 4}
Output: 0

A Naive 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.

An efficient approach (when elements are from small range) is to use sorting and binary search method. Firstly we will sort the array so that we would able to apply binary search on it. Since we need to maximize the value of arr[i] mod arr[j] so we iterate through each x(such x divisible by arr[j]) in range from 2*arr[j] to M+arr[j], where M is Maximum value of sequence. For each value of x we need to find maximum value of arr[i] such that arr[i] < x.
By doing this we would assure that we have chosen only those values of arr[i] that will give the maximum value of arr[i] mod arr[j]. After that we just need to repeat the above process for other values of arr[j] and update the answer by value a[i] mod arr[j]. For 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++

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// C++ program to find Maximum modulo value
#include <bits/stdc++.h>
using namespace std;
  
int maxModValue(int arr[], int n)
{
    int ans = 0;
    // Sort the array[] by using inbuilt sort function
    sort(arr, arr + n);
  
    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, arr + n, i) - arr;
  
            // Update the answer
            ans = max(ans, arr[ind - 1] % arr[j]);
        }
    }
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 3, 4, 5, 9, 11 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << maxModValue(arr, n);
}

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Java

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// Java program to find Maximum modulo value
  
import java.util.Arrays;
  
class Test {
    static int maxModValue(int arr[], int n)
    {
        int ans = 0;
  
        // Sort the array[] by using inbuilt sort function
        Arrays.sort(arr);
  
        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
  
                int ind = Arrays.binarySearch(arr, i);
  
                if (ind < 0)
                    ind = Math.abs(ind + 1);
  
                else {
                    while (arr[ind] == i) {
                        ind--;
  
                        if (ind == 0) {
                            ind = -1;
                            break;
                        }
                    }
                    ind++;
                }
  
                // Update the answer
                ans = Math.max(ans, arr[ind - 1] % arr[j]);
            }
        }
        return ans;
    }
  
    // Driver method
    public static void main(String args[])
    {
        int arr[] = { 3, 4, 5, 9, 11 };
        System.out.println(maxModValue(arr, arr.length));
    }
}

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Python3

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# Python3 program to find Maximum modulo value 
  
def maxModValue(arr, n): 
  
    ans = 0
      
    # Sort the array[] by using inbuilt
    # sort function 
    arr = sorted(arr) 
  
    for j in range(n - 2, -1, -1): 
          
        # 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
        i = 2 * arr[j]
        while(i <= arr[n - 1] + arr[j]):
              
            # Fetch the index which is greater than or 
            # equals to arr[i] by using binary search 
            # inbuilt lower_bound() function of C++ 
            ind = 0
            for k in arr:
                if k >= i:
                    ind = arr.index(k)
  
            # Update the answer 
            ans = max(ans, arr[ind - 1] % arr[j])
            i += arr[j]
  
    return ans 
  
# Driver Code
arr = [3, 4, 5, 9, 11 ]
n = 5
print(maxModValue(arr, n))
  
# This code is contributed by
# Shubham Singh(SHUBHAMSINGH10)

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C#

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// C# program to find Maximum modulo value
using System;
  
public class GFG {
      
    static int maxModValue(int[] arr, int n)
    {
          
        int ans = 0;
  
        // Sort the array[] by using inbuilt
        // sort function
        Array.Sort(arr);
  
        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
  
                int ind = Array.BinarySearch(arr, i);
  
                if (ind < 0)
                    ind = Math.Abs(ind + 1);
  
                else {
                    while (arr[ind] == i) {
                        ind--;
  
                        if (ind == 0) {
                            ind = -1;
                            break;
                        }
                    }
                    ind++;
                }
  
                // Update the answer
                ans = Math.Max(ans, arr[ind - 1]
                                       % arr[j]);
            }
        }
          
        return ans;
    }
  
    // Driver method
    public static void Main()
    {
          
        int[] arr = { 3, 4, 5, 9, 11 };
          
        Console.WriteLine(
                 maxModValue(arr, arr.Length));
    }
}
  
// This code is contributed by Sam007.

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Output:
4

Time complexity: O(nlog(n) + Mlog(M)) where n is total number of elements and M is maximum value of all the elements.
Auxiliary space: O(1)

This blog is contributed by Shubham Bansal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.



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