Minimum sum of two integers whose product is strictly greater than N

Given an integer N, the task is to find two integers with minimum possible sum such that their product is strictly greater than N.

Examples:

Input: N = 10
Output: 7
Explanation: The integers are 3 and 4. Their product is 3 × 4 = 12, which is greater than N.

Input: N = 1
Output: 3
Explanation: The integers are 1 and 2. Their product is 1 × 2 = 2, which is greater than N.

Naive Approach: Let the required numbers be A and B. The idea is based on the observation that in order to minimize their sum A should be the smallest number greater than √N. Once A is found, B will be equal to the smallest number for which A×B > N, which can be found linearly. 

Time Complexity: O(√N)
Auxiliary Space: O(1)

Efficient Approach: The above solution can be optimized by using Binary Search to find A and B. Follow the steps below to solve the problem:

• Initialize two variables low = 0  and high = 10⁹.
• Iterate until (high – low) is greater than 1 and do the following:
  • Find the value of middle-range mid as (low + high)/2.
  • Now, compare √N with the middle element mid, and if √N is less than or equal to the middle element,  then high as mid.
  • Else, update low as mid.
• After all the above steps set A = high.
• Repeat the same procedure to find B such that A×B > N.
• After the above steps, print the sum of A and B as the result.

Below is the implementation of the above approach:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
#define ll long long int

// Function to find the minimum sum of
// two integers such that their product
// is strictly greater than N
void minSum(int N)
{
    // Initialise low as 0 and
    // high as 1e9
    ll low = 0, high = 1e9;

    // Iterate to find the first number
    while (low + 1 < high) {

        // Find the middle value
        ll mid = low + (high - low) / 2;

        // If mid^2 is greater than
        // equal to A, then update
        // high to mid
        if (mid * mid >= N) {
            high = mid;
        }

        // Otherwise update low
        else {
            low = mid;
        }
    }

    // Store the first number
    ll first = high;

    // Again, set low as 0 and
    // high as 1e9
    low = 0;
    high = 1e9;

    // Iterate to find the second number
    while (low + 1 < high) {

        // Find the middle value
        ll mid = low + (high - low) / 2;

        // If first number * mid is
        // greater than N then update
        // high to mid
        if (first * mid > N) {
            high = mid;
        }

        // Else, update low to mid
        else {
            low = mid;
        }
    }

    // Store the second number
    ll second = high;

    // Print the result
    cout << first + second;
}

// Driver Code
int main()
{
    int N = 10;

    // Function Call
    minSum(N);

    return 0;
}

Java

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

class GFG{

// Function to find the minimum sum of
// two integers such that their product
// is strictly greater than N
static void minSum(int N)
{

    // Initialise low as 0 and
    // high as 1e9
    long low = 0, high = 1000000000;

    // Iterate to find the first number
    while (low + 1 < high)
    {

        // Find the middle value
        long mid = low + (high - low) / 2;

        // If mid^2 is greater than
        // equal to A, then update
        // high to mid
        if (mid * mid >= N)
        {
            high = mid;
        }

        // Otherwise update low
        else
        {
            low = mid;
        }
    }

    // Store the first number
    long first = high;

    // Again, set low as 0 and
    // high as 1e9
    low = 0;
    high = 1000000000;

    // Iterate to find the second number
    while (low + 1 < high)
    {

        // Find the middle value
        long mid = low + (high - low) / 2;

        // If first number * mid is
        // greater than N then update
        // high to mid
        if (first * mid > N)
        {
            high = mid;
        }

        // Else, update low to mid
        else
        {
            low = mid;
        }
    }

    // Store the second number
    long second = high;

    // Print the result
    System.out.println(first + second);
}

// Driver Code
public static void main(String[] args)
{
    int N = 10;

    // Function Call
    minSum(N);
}
}

// This code is contributed by Dharanendra L V

Python3

# Python3 program for the above approach

# Function to find the minimum sum of
# two integers such that their product
# is strictly greater than N
def minSum(N):

    # Initialise low as 0 and
    # high as 1e9
    low = 0
    high = 1000000000

    # Iterate to find the first number
    while (low + 1 < high):

        # Find the middle value
        mid = low + (high - low) / 2

        # If mid^2 is greater than
        # equal to A, then update
        # high to mid
        if (mid * mid >= N):
            high = mid

        # Otherwise update low
        else:
            low = mid

    # Store the first number
    first = high

    # Again, set low as 0 and
    # high as 1e9
    low = 0
    high = 1000000000

    # Iterate to find the second number
    while (low + 1 < high):

        # Find the middle value
        mid = low + (high - low) / 2

        # If first number * mid is
        # greater than N then update
        # high to mid
        if (first * mid > N):
            high = mid

        # Else, update low to mid
        else:
            low = mid

    # Store the second number
    second = high

    # Print the result
    print(round(first + second))

# Driver Code
N = 10

# Function Call
minSum(N)

# This code is contributed by Dharanendra L V

C#

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

class GFG{

// Function to find the minimum sum of
// two integers such that their product
// is strictly greater than N
static void minSum(int N)
{

    // Initialise low as 0 and
    // high as 1e9
    long low = 0, high = 1000000000;

    // Iterate to find the first number
    while (low + 1 < high)
    {

        // Find the middle value
        long mid = low + (high - low) / 2;

        // If mid^2 is greater than
        // equal to A, then update
        // high to mid
        if (mid * mid >= N)
        {
            high = mid;
        }

        // Otherwise update low
        else
        {
            low = mid;
        }
    }

    // Store the first number
    long first = high;

    // Again, set low as 0 and
    // high as 1e9
    low = 0;
    high = 1000000000;

    // Iterate to find the second number
    while (low + 1 < high)
    {

        // Find the middle value
        long mid = low + (high - low) / 2;

        // If first number * mid is
        // greater than N then update
        // high to mid
        if (first * mid > N)
        {
            high = mid;
        }

        // Else, update low to mid
        else
        {
            low = mid;
        }
    }

    // Store the second number
    long second = high;

    // Print the result
    Console.WriteLine( first + second);
}

// Driver Code
static public void Main()
{
    int N = 10;

    // Function Call
    minSum(N);
}
}

// This code is contributed by Dharanendra L V

Javascript

<script>
// Javascript program for the above approach

// Function to find the minimum sum of
// two integers such that their product
// is strictly greater than N
function minSum(N)
{
    // Initialise low as 0 and
    // high as 1e9
    let low = 0, high = 1000000000;

    // Iterate to find the first number
    while (low + 1 < high) {

        // Find the middle value
        let mid = low + parseInt((high - low) / 2);

        // If mid^2 is greater than
        // equal to A, then update
        // high to mid
        if (mid * mid >= N) {
            high = mid;
        }

        // Otherwise update low
        else {
            low = mid;
        }
    }

    // Store the first number
    let first = high;

    // Again, set low as 0 and
    // high as 1e9
    low = 0;
    high = 1000000000;

    // Iterate to find the second number
    while (low + 1 < high) {

        // Find the middle value
        let mid = low + parseInt((high - low) / 2);

        // If first number * mid is
        // greater than N then update
        // high to mid
        if (first * mid > N) {
            high = mid;
        }

        // Else, update low to mid
        else {
            low = mid;
        }
    }

    // Store the second number
    let second = high;

    // Print the result
    document.write(first + second);
}

// Driver Code
    let N = 10;

    // Function Call
    minSum(N);

// This code is contributed by rishavmahato348.
</script>

Output:

7

Time Complexity: O(log N)
Auxiliary Space: O(1)

Most Efficient Approach: To optimize the above approach, the idea is based on Inequality of Arithmetic and Geometric progression as illustrated below.

From the inequality, If there are two integers A and B, 
(A + B)/2 ≥ √(A×B)
Now, A×B = Product of the two integers, which is N and A+B is sum(=S).
Therefore, S ≥ 2*√N
To get strictly greater product than N, the above equation transforms to: S ≥ 2*√(N+1)

Below is the program for the above approach:

C++14

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the minimum sum of
// two integers such that their product
// is strictly greater than N
void minSum(int N)
{
    // Store the answer using the
    // AP-GP inequality
    int ans = ceil(2 * sqrt(N + 1));

    // Print the answer
    cout << ans;
}

// Driver Code
int main()
{
    int N = 10;

    // Function Call
    minSum(N);

    return 0;
}

Java

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

class GFG{


// Function to find the minimum sum of
// two integers such that their product
// is strictly greater than N
static void minSum(int N)
{

    // Store the answer using the
    // AP-GP inequality
    int ans = (int)Math.ceil(2 * Math.sqrt(N + 1));

    // Print the answer
    System.out.println( ans);
}

// Driver Code
public static void main(String[] args)
{
    int N = 10;

    // Function Call
    minSum(N);
}
}

// This code is contributed by Dharanendra L V

Python3

# Python3 program for the above approach
import math

# Function to find the minimum sum of
# two integers such that their product
# is strictly greater than N
def minSum(N):

    # Store the answer using the
    # AP-GP inequality
    ans = math.ceil(2 * math.sqrt(N + 1))

    # Print the result
    print(math.trunc(ans))

# Driver Code
N = 10

# Function Call
minSum(N)

# This code is contributed by Dharanendra L V

C#

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

class GFG{

// Function to find the minimum sum of
// two integers such that their product
// is strictly greater than N
static void minSum(int N)
{

    // Store the answer using the
    // AP-GP inequality
    int ans = (int)Math.Ceiling(2 * Math.Sqrt(N + 1));

    // Print the answer
    Console.WriteLine( ans);
}

// Driver Code
static public void Main()
{
    int N = 10;

    // Function Call
    minSum(N);
}
}

// This code is contributed by Dharanendra L V

Javascript

<script>

// Javascript program for the above approach

// Function to find the minimum sum of
// two integers such that their product
// is strictly greater than N
function minSum(N)
{
    // Store the answer using the
    // AP-GP inequality
    let ans = Math.ceil(2 * Math.sqrt(N + 1));

    // Print the answer
    document.write(ans);
}

// Driver Code
let N = 10;

// Function Call
minSum(N);

</script>

Output:

7

Time Complexity: O(logN) because it is using inbuilt sqrt function
Auxiliary Space: O(1)