来自数组的总和为 K 的最小子数组

给定一个由N 个整数组成的数组arr[] ，任务是找到总和等于K的最小子数组的长度。

例子：

Input: arr[] = {2, 4, 6, 10, 2, 1}, K = 12 
Output: 2 
Explanation: 
All possible subarrays with sum 12 are {2, 4, 6} and {10, 2}.

Input: arr[] = { 1, 2, 4, 3, 2, 4, 1 }, K = 7 
Output: 2

朴素方法：解决问题的最简单方法是生成所有可能的子数组，并为每个子数组检查其总和是否等于K。打印所有此类子数组的最小长度。

时间复杂度： O(N ² )
辅助空间： O(1)

高效的方法：可以使用 Prefix Sum 技术和 HashMap 进一步优化上述方法。请按照以下步骤解决问题：

1. 计算每个索引的前缀总和并将(索引，前缀总和)存储为映射中的键值对。
2. 遍历前缀和数组并计算前缀和与所需和之间的差值。
3. 如果 HashMap 中存在差值，则表示存在一个总和等于K的子数组，然后将子数组的长度与得到的最小长度进行比较，并相应地更新最小长度。

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 with sum K
int subArraylen(int arr[], int n, int K)
{
    // Stores the frequency of
    // prefix sums in the array
    unordered_map mp;
 
    mp[arr[0]] = 0;
 
    for (int i = 1; i < n; i++) {
 
        arr[i] = arr[i] + arr[i - 1];
        mp[arr[i]] = i;
    }
 
    // Initialize len as INT_MAX
    int len = INT_MAX;
 
    for (int i = 0; i < n; i++) {
 
        // If sum of array till i-th
        // index is less than K
        if (arr[i] < K)
 
            // No possible subarray
            // exists till i-th index
            continue;
 
        else {
 
            // Find the exceeded value
            int x = arr[i] - K;
 
            // If exceeded value is zero
            if (x == 0)
                len = min(len, i);
 
            if (mp.find(x) == mp.end())
                continue;
            else {
                len = min(len, i - mp[x]);
            }
        }
    }
 
    return len;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 4, 3, 2, 4, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    int K = 7;
 
    int len = subArraylen(arr, n, K);
 
    if (len == INT_MAX) {
        cout << "-1";
    }
    else {
        cout << len << endl;
    }
 
    return 0;
}

Java

// Java Program to implement
// the above approach
import java.util.*;
class GFG{
 
// Function to find the length of the
// smallest subarray with sum K
static int subArraylen(int arr[], int n, int K)
{
    // Stores the frequency of
    // prefix sums in the array
    HashMap mp = new HashMap();
 
    mp.put(arr[0], 0);
 
    for (int i = 1; i < n; i++)
    {
        arr[i] = arr[i] + arr[i - 1];
        mp.put(arr[i], i);
    }
 
    // Initialize len as Integer.MAX_VALUE
    int len = Integer.MAX_VALUE;
 
    for (int i = 0; i < n; i++)
    {
 
        // If sum of array till i-th
        // index is less than K
        if (arr[i] < K)
 
            // No possible subarray
            // exists till i-th index
            continue;
        else
        {
 
            // Find the exceeded value
            int x = K - arr[i];
 
            // If exceeded value is zero
            if (x == 0)
                len = Math.min(len, i);
 
            if (mp.containsValue(x))
                continue;
            else
            {
                len = Math.min(len, i );
            }
        }
    }
    return len;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 4, 3, 2, 4, 1 };
    int n = arr.length;
 
    int K = 7;
 
    int len = subArraylen(arr, n, K);
 
    if (len == Integer.MAX_VALUE)
    {
        System.out.print("-1");
    }
    else
    {
        System.out.print(len + "\n");
    }
}
}
 
// This code is contributed by Rohit_ranjan

Python3

# Python3 program to implement
# the above approach
from collections import defaultdict
import sys
 
# Function to find the length of the
# smallest subarray with sum K
def subArraylen(arr, n, K):
 
    # Stores the frequency of
    # prefix sums in the array
    mp = defaultdict(lambda : 0)
 
    mp[arr[0]] = 0
 
    for i in range(1, n):
        arr[i] = arr[i] + arr[i - 1]
        mp[arr[i]] = i
 
    # Initialize ln
    ln = sys.maxsize
 
    for i in range(n):
 
        # If sum of array till i-th
        # index is less than K
        if(arr[i] < K):
 
            # No possible subarray
            # exists till i-th index
            continue
        else:
             
            # Find the exceeded value
            x = K - arr[i]
 
            # If exceeded value is zero
            if(x == 0):
                ln = min(ln, i)
 
            if(x in mp.keys()):
                continue
            else:
                ln = min(ln, i - mp[x])
 
    return ln
 
# Driver Code
arr = [ 1, 2, 4, 3, 2, 4, 1 ]
n = len(arr)
 
K = 7
 
ln = subArraylen(arr, n, K)
 
# Function call
if(ln == sys.maxsize):
    print("-1")
else:
    print(ln)
 
# This code is contributed by Shivam Singh

C#

// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to find the length of the
// smallest subarray with sum K
static int subArraylen(int []arr, int n, int K)
{
    // Stores the frequency of
    // prefix sums in the array
    Dictionary mp = new Dictionary();
 
    mp.Add(arr[0], 0);
 
    for (int i = 1; i < n; i++)
    {
        arr[i] = arr[i] + arr[i - 1];
        mp.Add(arr[i], i);
    }
 
    // Initialize len as int.MaxValue
    int len = int.MaxValue;
 
    for (int i = 0; i < n; i++)
    {
 
        // If sum of array till i-th
        // index is less than K
        if (arr[i] < K)
 
            // No possible subarray
            // exists till i-th index
            continue;
        else
        {
 
            // Find the exceeded value
            int x = K - arr[i];
 
            // If exceeded value is zero
            if (x == 0)
                len = Math.Min(len, i);
 
            if (mp.ContainsValue(x))
                continue;
            else
            {
                len = Math.Min(len, i );
            }
        }
    }
    return len;
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 4, 3, 2, 4, 1 };
    int n = arr.Length;
 
    int K = 7;
 
    int len = subArraylen(arr, n, K);
 
    if (len == int.MaxValue)
    {
        Console.Write("-1");
    }
    else
    {
        Console.Write(len + "\n");
    }
}
}
 
// This code is contributed by Rohit_ranjan

Javascript

 
// JavaScript Program to implement
// the above approach
 
// Function to find the length of the
// smallest subarray with sum K
function subArraylen(arr, n, K)
{
    // Stores the frequency of
    // prefix sums in the array
    var mp = new Map();
 
    mp.set(arr[0],0);
 
    for (var i = 1; i < n; i++) {
 
        arr[i] = arr[i] + arr[i - 1];
        mp.set(arr[i],i);
    }
 
    // Initialize len as INT_MAX
    var len = 10000000000;
 
    for (var i = 0; i < n; i++) {
 
        // If sum of array till i-th
        // index is less than K
        if (arr[i] < K)
 
            // No possible subarray
            // exists till i-th index
            continue;
 
        else {
 
            // Find the exceeded value
            var x = arr[i] - K;
 
            // If exceeded value is zero
            if (x == 0)
                len = Math.min(len, i);
 
            if (!mp.has(x))
                continue;
            else {
                len = Math.min(len, i - mp.get(x));
            }
        }
    }
 
    return len;
}
 
// Driver Code
var arr = [1, 2, 4, 3, 2, 4, 1];
var n = arr.length;
var K = 7;
var len = subArraylen(arr, n, K);
if (len == 1000000000) {
    document.write( "-1");
}
else {
    document.write( len );
}
 
 

输出

2

时间复杂度： O(NlogN)
辅助空间： O(N)

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