大小大于K且总和大于给定值的最小子数组

给定一个大小为N的数组arr [] ，两个正整数K和S ，任务是查找大小大于K且其总和大于S的最小子数组的长度。

例子：

Input: arr[] = {1, 2, 3, 4, 5}, K = 1, S = 8
Output: 2
Explanation: 
Smallest subarray with sum greater than S(=8) is {4, 5}
Therefore, the required output is 2.

Input: arr[] = {1, 3, 5, 1, 8, 2, 4}, K= 2, S= 13
Output: 3

方法：可以使用滑动窗口技术解决该问题。请按照以下步骤解决问题：

1. 初始化两个变量，例如start = 0和end = 0 ，分别存储当前子数组的第一个和最后一个索引。
2. 遍历数组arr []并通过增加end并将arr [end]添加到当前子数组的总和中来。
3. 如果当前子数组的所有元素的总和小于或等于S或当前子数组的长度小于或等于K ，则增加当前子数组的长度( end ++ )。
4. 否则，通过删除当前子数组的第一个元素来减少当前子数组的长度(start- = 1 )。通过将其与当前子数组的长度进行比较，来更新到目前为止找到的所需子数组的最小长度。
5. 最后，打印满足条件的所需子数组的最小长度。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to find the length of the
// smallest subarray of size > K with
// sum greater than S
int smallestSubarray(int K, int S,
                     int arr[], int N)
{
    // Store the first index of
    // the current subarray
    int start = 0;
 
    // Store the last index of
    // the the current subarray
    int end = 0;
 
    // Store the sum of the
    // current subarray
    int currSum = arr[0];
 
    // Store the length of
    // the smallest subarray
    int res = INT_MAX;
 
    while (end < N - 1) {
 
        // If sum of the current subarray <= S
        // or length of current subarray <= K
        if (currSum <= S
            || (end - start + 1) <= K) {
            // Increase the subarray
            // sum and size
            currSum += arr[++end];
        }
 
        // Otherwise
        else {
 
            // Update to store the minimum
            // size of subarray obtained
            res = min(res, end - start + 1);
 
            // Decrement current subarray
            // size by removing first element
            currSum -= arr[start++];
        }
    }
 
    // Check if it is possible to reduce
    // the length of the current window
    while (start < N) {
        if (currSum > S
            && (end - start + 1) > K)
            res = min(res, (end - start + 1));
 
        currSum -= arr[start++];
    }
    return res;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int K = 1, S = 8;
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << smallestSubarray(K, S, arr, N);
}

Java

// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
 
// Function to find the length of the
// smallest subarray of size > K with
// sum greater than S
public static int smallestSubarray(int K, int S,
                                   int[] arr, int N)
{
     
    // Store the first index of
    // the current subarray
    int start = 0;
 
    // Store the last index of
    // the the current subarray
    int end = 0;
 
    // Store the sum of the
    // current subarray
    int currSum = arr[0];
 
    // Store the length of
    // the smallest subarray
    int res = Integer.MAX_VALUE;
 
    while (end < N - 1)
    {
         
        // If sum of the current subarray <= S
        // or length of current subarray <= K
        if (currSum <= S ||
           (end - start + 1) <= K)
        {
             
            // Increase the subarray
            // sum and size
            currSum += arr[++end];
        }
 
        // Otherwise
        else
        {
 
            // Update to store the minimum
            // size of subarray obtained
            res = Math.min(res, end - start + 1);
 
            // Decrement current subarray
            // size by removing first element
            currSum -= arr[start++];
        }
    }
 
    // Check if it is possible to reduce
    // the length of the current window
    while (start < N)
    {
        if (currSum > S && (end - start + 1) > K)
            res = Math.min(res, (end - start + 1));
 
        currSum -= arr[start++];
    }
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
    int[] arr = { 1, 2, 3, 4, 5 };
    int K = 1, S = 8;
    int N = arr.length;
     
    System.out.print(smallestSubarray(K, S, arr, N));
}
}
 
// This code is contributed by akhilsaini

Python3

# Python3 program to implement
# the above approach
import sys
 
# Function to find the length of the
# smallest subarray of size > K with
# sum greater than S
def smallestSubarray(K, S, arr, N):
   
  # Store the first index of
  # the current subarray
  start = 0
 
  # Store the last index of
  # the the current subarray
  end = 0
 
  # Store the sum of the
  # current subarray
  currSum = arr[0]
 
  # Store the length of
  # the smallest subarray
  res = sys.maxsize
 
  while end < N - 1:
 
      # If sum of the current subarray <= S
      # or length of current subarray <= K
      if ((currSum <= S) or
         ((end - start + 1) <= K)):
           
          # Increase the subarray
          # sum and size
          end = end + 1;
          currSum += arr[end]
 
      # Otherwise
      else:
 
          # Update to store the minimum
          # size of subarray obtained
          res = min(res, end - start + 1)
 
          # Decrement current subarray
          # size by removing first element
          currSum -= arr[start]
          start = start + 1
 
  # Check if it is possible to reduce
  # the length of the current window
  while start < N:
      if ((currSum > S) and
         ((end - start + 1) > K)):
          res = min(res, (end - start + 1))
       
      currSum -= arr[start]
      start = start + 1
 
  return res;
 
# Driver Code
if __name__ == "__main__":
     
  arr = [ 1, 2, 3, 4, 5 ]
  K = 1
  S = 8
  N = len(arr)
   
  print(smallestSubarray(K, S, arr, N))
 
# This code is contributed by akhilsaini

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Function to find the length of the
// smallest subarray of size > K with
// sum greater than S
static int smallestSubarray(int K, int S,
                            int[] arr, int N)
{
     
    // Store the first index of
    // the current subarray
    int start = 0;
 
    // Store the last index of
    // the the current subarray
    int end = 0;
 
    // Store the sum of the
    // current subarray
    int currSum = arr[0];
 
    // Store the length of
    // the smallest subarray
    int res = int.MaxValue;
 
    while (end < N - 1)
    {
         
        // If sum of the current subarray <= S
        // or length of current subarray <= K
        if (currSum <= S ||
           (end - start + 1) <= K)
        {
             
            // Increase the subarray
            // sum and size
            currSum += arr[++end];
        }
 
        // Otherwise
        else
        {
 
            // Update to store the minimum
            // size of subarray obtained
            res = Math.Min(res, end - start + 1);
 
            // Decrement current subarray
            // size by removing first element
            currSum -= arr[start++];
        }
    }
 
    // Check if it is possible to reduce
    // the length of the current window
    while (start < N)
    {
        if (currSum > S && (end - start + 1) > K)
            res = Math.Min(res, (end - start + 1));
 
        currSum -= arr[start++];
    }
    return res;
}
 
// Driver Code
static public void Main()
{
    int[] arr = { 1, 2, 3, 4, 5 };
    int K = 1, S = 8;
    int N = arr.Length;
     
    Console.Write(smallestSubarray(K, S, arr, N));
}
}
 
// This code is contributed by akhilsaini

Javascript

// JavaScript program to implement
// the above approach
 
// Function to find the length of the
// smallest subarray of size > K with
// sum greater than S
function smallestSubarray(K, S, arr, N)
{
 
    // Store the first index of
    // the current subarray
    let start = 0;
 
    // Store the last index of
    // the the current subarray
    let end = 0;
 
    // Store the sum of the
    // current subarray
    let currSum = arr[0];
 
    // Store the length of
    // the smallest subarray
    let res = Number.MAX_SAFE_INTEGER;
    while (end < N - 1)
    {
 
        // If sum of the current subarray <= S
        // or length of current subarray <= K
        if (currSum <= S
            || (end - start + 1) <= K)
        {
         
            // Increase the subarray
            // sum and size
            currSum += arr[++end];
        }
 
        // Otherwise
        else {
 
            // Update to store the minimum
            // size of subarray obtained
            res = Math.min(res, end - start + 1);
 
            // Decrement current subarray
            // size by removing first element
            currSum -= arr[start++];
        }
    }
 
    // Check if it is possible to reduce
    // the length of the current window
    while (start < N)
    {
        if (currSum > S
            && (end - start + 1) > K)
            res = Math.min(res, (end - start + 1));
 
        currSum -= arr[start++];
    }
    return res;
}
 
// Driver Code
    let arr = [ 1, 2, 3, 4, 5 ];
    let K = 1, S = 8;
    let N = arr.length;
    document.write(smallestSubarray(K, S, arr, N));
 
// This code is contributed by Surbhi tyagi.

输出：

2

时间复杂度： O(N)
辅助空间： O(1)