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Minimum sum of two integers whose product is strictly greater than N
  • Last Updated : 08 Jan, 2021

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

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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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

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output: 

7

 

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

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