Length of Smallest subarray in range 1 to N with sum greater than a given value

Difficulty Level : Easy
Last Updated : 09 May, 2020

Given two numbers N and S, the task is to find the length of smallest subarray in range (1, N) such that the sum of those chosen numbers is greater than S.

Examples:

Input: N = 5, S = 11
Output: 3
Explanation:
Smallest subarray with sum > 11 = {5, 4, 3}

Input: N = 4, S = 7
Output: 3
Explanation:
Smallest subarray with sum > 7 = {4, 3, 2}

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Naive Approach: A brute force method is to select elements in reverse order until the sum of all the selected elements is less than or equal to the given number.

Below is the implementation of the above approach.

C++

// C++ implementation of the above implementation 
  
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to return the count 
// of minimum elements such that  
// the sum of those elements is > S. 
int countNumber(int N, int S) 
{ 
    int countElements = 0; 
    // Initialize currentSum = 0 
  
    int currSum = 0; 
  
    // Loop from N to 1 to add the numbers 
    // and check the condition. 
    while (currSum <= S) { 
        currSum += N; 
        N--; 
        countElements++; 
    } 
  
    return countElements; 
} 
  
// Driver code 
int main() 
{ 
    int N, S; 
    N = 5; 
    S = 11; 
  
    int count = countNumber(N, S); 
  
    cout << count << endl; 
  
    return 0; 
} 

Java

// Java implementation of the above implementation  
class GFG  
{ 
  
    // Function to return the count  
    // of minimum elements such that  
    // the sum of those elements is > S.  
    static int countNumber(int N, int S)  
    {  
        int countElements = 0;  
  
        // Initialize currentSum = 0  
        int currSum = 0;  
      
        // Loop from N to 1 to add the numbers  
        // and check the condition.  
        while (currSum <= S)  
        {  
            currSum += N;  
            N--;  
            countElements++;  
        }  
        return countElements;  
    }  
      
    // Driver code  
    public static void main (String[] args) 
    {  
        int N, S;  
        N = 5;  
        S = 11;  
      
        int count = countNumber(N, S);  
      
        System.out.println(count);  
    }  
} 
  
// This code is contributed by AnkitRai01 

Python

# Python implementation of the above implementation 
  
# Function to return the count 
# of minimum elements such that  
# the sum of those elements is > S. 
def countNumber(N, S): 
  
    countElements = 0; 
    currentSum = 0
  
    currSum = 0; 
  
    # Loop from N to 1 to add the numbers 
    # and check the condition. 
    while (currSum <= S) : 
        currSum += N; 
        N = N - 1; 
        countElements=countElements + 1; 
      
    return countElements; 
  
# Driver code 
N = 5; 
S = 11; 
count = countNumber(N, S); 
print(count) ; 
  
# This code is contributed by Shivi_Aggarwal 

C#

// C# implementation of the above implementation  
using System; 
  
class GFG  
{ 
  
    // Function to return the count  
    // of minimum elements such that  
    // the sum of those elements is > S.  
    static int countNumber(int N, int S)  
    {  
        int countElements = 0;  
  
        // Initialize currentSum = 0  
        int currSum = 0;  
      
        // Loop from N to 1 to add the numbers  
        // and check the condition.  
        while (currSum <= S)  
        {  
            currSum += N;  
            N--;  
            countElements++;  
        }  
        return countElements;  
    }  
      
    // Driver code  
    public static void Main() 
    {  
        int N, S;  
        N = 5;  
        S = 11;  
      
        int count = countNumber(N, S);  
      
        Console.WriteLine(count);  
    }  
} 
  
// This code is contributed by AnkitRai01 

Output:

Time Complexity: O(N)

Efficient Approach: The idea is to use Binary Search concept to solve the problem.

From the binary search concept, it is known that the concept can be applied when it is known that there is an order in the problem. That is, for every iteration, if it can be differentiated for sure that the required answer either lies in the first half or second half (i.e), there exists a pattern in the problem.

Therefore, binary search can be applied for the range in the following way:

Initialize start = 1 and end = N.
Find mid = start + (end - start) / 2.
If the sum of all the elements from the last element to mid element is less than or equal to the given sum, then end = mid else start = mid + 1.
Repeat step 2 while start is less than the end.

Below is the implementation of the above approach.

C++

// C++ implementation of the above approach. 
  
#include <iostream> 
using namespace std; 
  
// Function to do a binary search 
// on a given range. 
int usingBinarySearch(int start, int end, 
                      int N, int S) 
{ 
    if (start >= end) 
        return start; 
    int mid = start + (end - start) / 2; 
  
    // Total sum is the sum of N numbers. 
    int totalSum = (N * (N + 1)) / 2; 
  
    // Sum until mid 
    int midSum = (mid * (mid + 1)) / 2; 
  
    // If remaining sum is < the required value, 
    // then the required number is in the right half 
    if ((totalSum - midSum) <= S) { 
  
        return usingBinarySearch(start, mid, N, S); 
    } 
    return usingBinarySearch(mid + 1, end, N, S); 
} 
  
// Driver code 
int main() 
{ 
    int N, S; 
  
    N = 5; 
    S = 11; 
  
    cout << (N - usingBinarySearch(1, N, N, S) + 1) 
         << endl; 
  
    return 0; 
} 

Java

// Java implementation of the above approach.  
class GFG  
{ 
      
    // Function to do a binary search  
    // on a given range.  
    static int usingBinarySearch(int start, int end,  
                                int N, int S)  
    {  
        if (start >= end)  
            return start;  
        int mid = start + (end - start) / 2;  
      
        // Total sum is the sum of N numbers.  
        int totalSum = (N * (N + 1)) / 2;  
      
        // Sum until mid  
        int midSum = (mid * (mid + 1)) / 2;  
      
        // If remaining sum is < the required value,  
        // then the required number is in the right half  
        if ((totalSum - midSum) <= S) 
        {  
      
            return usingBinarySearch(start, mid, N, S);  
        }  
        return usingBinarySearch(mid + 1, end, N, S);  
    }  
      
    // Driver code  
    public static void main (String[] args) 
    {  
        int N, S;  
      
        N = 5;  
        S = 11;  
      
        System.out.println(N - usingBinarySearch(1, N, N, S) + 1) ; 
    } 
} 
  
// This code is contributed by AnkitRai01 

Python3

# Python3 implementation of the above approach.  
  
# Function to do a binary search  
# on a given range.  
def usingBinarySearch(start, end, N, S) :  
  
    if (start >= end) : 
        return start;  
          
    mid = start + (end - start) // 2;  
  
    # Total sum is the sum of N numbers.  
    totalSum = (N * (N + 1)) // 2;  
  
    # Sum until mid  
    midSum = (mid * (mid + 1)) // 2;  
  
    # If remaining sum is < the required value,  
    # then the required number is in the right half  
    if ((totalSum - midSum) <= S) : 
  
        return usingBinarySearch(start, mid, N, S);  
      
    return usingBinarySearch(mid + 1, end, N, S);  
  
# Driver code  
if __name__ == "__main__" :  
  
    N = 5;  
    S = 11;  
  
    print(N - usingBinarySearch(1, N, N, S) + 1) ;  
      
# This code is contributed by AnkitRai01 

C#

// C# implementation of the above approach.  
using System; 
  
class GFG  
{ 
      
    // Function to do a binary search  
    // on a given range.  
    static int usingBinarySearch(int start, int end,  
                                int N, int S)  
    {  
        if (start >= end)  
            return start;  
        int mid = start + (end - start) / 2;  
      
        // Total sum is the sum of N numbers.  
        int totalSum = (N * (N + 1)) / 2;  
      
        // Sum until mid  
        int midSum = (mid * (mid + 1)) / 2;  
      
        // If remaining sum is < the required value,  
        // then the required number is in the right half  
        if ((totalSum - midSum) <= S) 
        {  
      
            return usingBinarySearch(start, mid, N, S);  
        }  
        return usingBinarySearch(mid + 1, end, N, S);  
    }  
      
    // Driver code  
    public static void Main() 
    {  
        int N, S;  
      
        N = 5;  
        S = 11;  
      
        Console.WriteLine(N - usingBinarySearch(1, N, N, S) + 1) ; 
    } 
} 
  
// This code is contributed by AnkitRai01 

Output:

Time Complexity: O(log N)

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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