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 = 11Output:3Explanation:

Smallest subarray with sum > 11 = {5, 4, 3}

Input:N = 4, S = 7Output:3Explanation:

Smallest subarray with sum > 7 = {4, 3, 2}

**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` |

## Javascript

`<script>` `// JavaScript implementation of the above implementation` `// Function to return the count` `// of minimum elements such that` `// the sum of those elements is > S.` `function` `countNumber(N, S)` `{` ` ` `let countElements = 0;` ` ` ` ` `// Initialize currentSum = 0` ` ` `let 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` ` ` `let N, S;` ` ` `N = 5;` ` ` `S = 11;` ` ` `let count = countNumber(N, S);` ` ` `document.write(count + ` `"<br>"` `);` `// This code is contributed by Surbhi Tyagi.` `</script>` |

**Output:**

3

**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` |

## Javascript

`<script>` `// Javascript implementation of the above approach.` `// Function to do a binary search` `// on a given range.` `function` `usingBinarySearch(start, end, N, S)` `{` ` ` `if` `(start >= end)` ` ` `return` `start;` ` ` ` ` `let mid = start + (end - start) / 2;` ` ` `// Total sum is the sum of N numbers.` ` ` `let totalSum = (N * (N + 1)) / 2;` ` ` `// Sum until mid` ` ` `let 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` `let N, S;` `N = 5;` `S = 11;` `document.write((N - usingBinarySearch(` ` ` `1, N, N, S) + 1) + ` `"<br>"` `);` `// This code is contributed by Mayank Tyagi` `</script>` |

**Output:**

3

**Time Complexity: **O(log N)

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