Minimum product modulo N possible for any pair from a given range

Given three integers L, R, and N, the task is to find the minimum possible value of (i * j) % N, where L ? i < j ? R.

Examples:

Input: L = 2020, R = 2040, N = 2019
Output: 2
Explanation: (2020 * 2021) % 2019 = 2

Input: L = 15, R = 30, N = 15
Output: 0
Explanation: If one of the elements of the pair is 15, then the product of all such pairs will be divisible by 15. Therefore, the remainder will be 0, which is minimum possible.

Approach: The given problem can be solved by finding the difference between L and R. If the difference is at least N, then the result will be 0. Otherwise, iterate over the range [L, R] and find the minimum product.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

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

// Function to return the minimum
// possible value of (i * j) % N
void minModulo(int L, int R, int N)
{
    if (R - L < N) {

        // Stores the minimum remainder
        int ans = INT_MAX;

        // Iterate from L to R
        for (ll i = L; i <= R; i++)

            // Iterate from L to R
            for (ll j = L; j <= R; j++)
                if (i != j)
                    ans = min(0ll + ans,
                              (i * j) % N);

        // Print the minimum value
        // of remainder
        cout << ans;
    }

    // If R - L >= N
    else {
        cout << 0;
    }
}

// Driver Code
int main()
{
    int L = 6, R = 10, N = 2019;
    minModulo(L, R, N);

    return 0;
}

Java

// java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;

public class GFG
{

    // Function to return the minimum
    // possible value of (i * j) % N
    static void minModulo(int L, int R, int N)
    {
        if (R - L < N)
        {

            // Stores the minimum remainder
            int ans = Integer.MAX_VALUE;

            // Iterate from L to R
            for (int i = L; i <= R; i++)

                // Iterate from L to R
                for (int j = L; j <= R; j++)
                    if (i != j)
                        ans = Math.min(ans, (i * j) % N);

            // Print the minimum value
            // of remainder
            System.out.println(ans);
        }

        // If R - L >= N
        else {
            System.out.println(0);
        }
    }

    // Driver Code
    public static void main(String[] args)
    {

        int L = 6, R = 10, N = 2019;
        minModulo(L, R, N);
    }
}

// This code is contributed by Kingash.

Python3

# Python3 program for the above approach

# Function to return the minimum
# possible value of (i * j) % N
def minModulo(L, R, N):

    if (R - L < N):

        # Stores the minimum remainder
        ans = 10**9

        # Iterate from L to R
        for i in range(L, R + 1):

            # Iterate from L to R
            for j in range(L, R + 1):
                if (i != j):
                    ans = min(ans, (i * j) % N)

        # Print the minimum value
        # of remainder
        print (ans)

    # If R - L >= N
    else:
        print (0)

# Driver Code
if __name__ == '__main__':

    L, R, N = 6, 10, 2019
    minModulo(L, R, N)

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;

class GFG{

// Function to return the minimum
// possible value of (i * j) % N
static void minModulo(int L, int R, int N)
{
    if (R - L < N)
    {

        // Stores the minimum remainder
        int ans = Int32.MaxValue;

        // Iterate from L to R
        for(int i = L; i <= R; i++)

            // Iterate from L to R
            for(int j = L; j <= R; j++)
                if (i != j)
                    ans = Math.Min(ans, (i * j) % N);

        // Print the minimum value
        // of remainder
        Console.WriteLine(ans);
    }

    // If R - L >= N
    else
    {
        Console.WriteLine(0);
    }
}

// Driver Code
public static void Main(string[] args)
{
    int L = 6, R = 10, N = 2019;
    minModulo(L, R, N);
}
}

// This code is contributed by ukasp

Javascript

<script>

// Javascript implementation
// for the above approach

    // Function to return the minimum
    // possible value of (i * j) % N
    function minModulo(L, R, N)
    {
        if (R - L < N)
        {

            // Stores the minimum remainder
            let ans = Number.MAX_VALUE;

            // Iterate from L to R
            for (let i = L; i <= R; i++)

                // Iterate from L to R
                for (let j = L; j <= R; j++)
                    if (i != j)
                        ans = Math.min(ans, (i * j) % N);

            // Print the minimum value
            // of remainder
            document.write(ans);
        }

        // If R - L >= N
        else {
            document.write(0);
        }
    }


// Driver Code

    let L = 6, R = 10, N = 2019;
        minModulo(L, R, N);

    // This code is contributed by susmitakundugoaldanga.
</script>

Output:

42

Time Complexity: O((R – L)²)
Auxiliary Space: O(1), since no extra space has been taken.