Skip to content
Related Articles

Related Articles

Improve Article

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

  • Last Updated : 23 Apr, 2021
Geek Week

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(1)
Auxiliary Space: O(1)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :