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
天真的方法：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：
- 计算每个索引的前缀总和，并将其(索引，前缀总和)存储为映射中的键值对。
- 遍历前缀总和数组，并计算前缀总和与所需总和之间的差。
- 如果HashMap中存在差异值，则意味着存在一个总和等于K的子数组，然后将该子数组的长度与获得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 SinghC#
// 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 输出
2
时间复杂度： O(NlogN)
辅助空间： O(N)