给定由N个整数和整数K组成的数组arr [] ，任务是通过从数组的任一端删除数组元素并将其从K中减去来将K减小为0 。如果不可能将K减小为0 ，则打印“ -1” 。否则，请打印所需最少数量的此类操作。
例子：
Input: arr[] = {1, 3, 1, 1, 2}, K = 4
Output: 2
Explanation:
The given array is {1, 3, 1, 1, 2}
Operation1: Removing arr[0] (= 1) modifies arr[] to {3, 1, 1, 2}. Subtracting arr[0] from K reduces K to 4 – 1 = 3.
Operation2: Removing arr[0] (= 1) modifies arr[] to {1, 1, 2}. Subtracting arr[0] from K reduces K to 3 – 3 = 0.
Therefore, the total number of steps required is 2.
Input: arr[] = {1, 1, 3, 4}, K = 3
Output: -1
方法：可以根据以下观察结果解决给定问题：
- 考虑到需要分别从给定数组的前面和后面删除X和Y元素，以将K减小为0 。
- 因此，其余数组元素的总和必须等于(数组元素的总和– K) 。
因此，从上面的观察中，我们的想法是找到总和等于(数组元素之和– K)的子数组的最大长度(例如L ) 。因此，将(N – L)的值打印为要删除的最小元素，以将K减小为0 。如果不存在任何此类子数组，则打印“ -1” 。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the length of
// longest subarray having sum K
int longestSubarray(int arr[],
int N, int K)
{
// Stores the index of
// the prefix sum
unordered_map um;
int sum = 0, maxLen = 0;
// Traverse the given array
for (int i = 0; i < N; i++) {
// Update sum
sum += arr[i];
// If the subarray starts
// from index 0
if (sum == K)
maxLen = i + 1;
// Add the current prefix sum
// with index if it is not
// present in the map um
if (um.find(sum) == um.end())
um[sum] = i;
// Check if sum - K is
// present in Map um or not
if (um.find(sum - K) != um.end()) {
// Update the maxLength
if (maxLen < (i - um[sum - K]))
maxLen = i - um[sum - K];
}
}
// Return the required maximum length
return maxLen;
}
// Function to find the minimum removal of
// array elements required to reduce K to 0
void minRequiredOperation(int arr[],
int N, int K)
{
// Stores the sum of the array
int TotalSum = 0;
// Traverse the array arr[]
for (int i = 0; i < N; i++)
// Update sum of the array
TotalSum += arr[i];
// Find maxLen
int maxLen = longestSubarray(
arr, N, TotalSum - K);
// If the subarray with
// sum doesn't exist
if (maxLen == -1) {
cout << -1;
}
// Otherwise, print the
// required maxmimum length
else
cout << N - maxLen;
}
// Driver Code
int main()
{
int arr[] = { 1, 3, 1, 1, 2 };
int K = 4;
int N = sizeof(arr) / sizeof(arr[0]);
minRequiredOperation(arr, N, K);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the length of
// longest subarray having sum K
static int longestSubarray(int[] arr, int N, int K)
{
// Stores the index of
// the prefix sum
HashMap um = new HashMap();
int sum = 0, maxLen = 0;
// Traverse the given array
for(int i = 0; i < N; i++)
{
// Update sum
sum += arr[i];
// If the subarray starts
// from index 0
if (sum == K)
maxLen = i + 1;
// Add the current prefix sum
// with index if it is not
// present in the map um
if (!um.containsKey(sum))
um.put(sum, i);
// Check if sum - K is
// present in Map um or not
if (um.containsKey(sum - K))
{
// Update the maxLength
if (maxLen < (i - um.get(sum - K)))
maxLen = i - um.get(sum - K);
}
}
// Return the required maximum length
return maxLen;
}
// Function to find the minimum removal of
// array elements required to reduce K to 0
static void minRequiredOperation(int[] arr, int N,
int K)
{
// Stores the sum of the array
int TotalSum = 0;
// Traverse the array arr[]
for(int i = 0; i < N; i++)
// Update sum of the array
TotalSum += arr[i];
// Find maxLen
int maxLen = longestSubarray(arr, N, TotalSum - K);
// If the subarray with
// sum doesn't exist
if (maxLen == -1)
{
System.out.println(-1);
}
// Otherwise, print the
// required maxmimum length
else
System.out.println(N - maxLen);
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 1, 3, 1, 1, 2 };
int K = 4;
int N = arr.length;
minRequiredOperation(arr, N, K);
}
}
// This code is contributed by ukasp Python3
# Python3 program for the above approach
# Function to find the length of
# longest subarray having sum K
def longestSubarray(arr, N, K):
# Stores the index of
# the prefix sum
um = {}
sum , maxLen = 0, 0
# Traverse the given array
for i in range(N):
# Update sum
sum += arr[i]
# If the subarray starts
# from index 0
if (sum == K):
maxLen = i + 1
# Add the current prefix sum
# with index if it is not
# present in the map um
if (sum not in um):
um[sum] = i
# Check if sum - K is
# present in Map um or not
if ((sum - K) in um):
# Update the maxLength
if (maxLen < (i - um[sum - K])):
maxLen = i - um[sum - K]
# Return the required maximum length
return maxLen
# Function to find the minimum removal of
# array elements required to reduce K to 0
def minRequiredOperation(arr, N, K):
# Stores the sum of the array
TotalSum = 0
# Traverse the array arr[]
for i in range(N):
# Update sum of the array
TotalSum += arr[i]
# Find maxLen
maxLen = longestSubarray(arr, N, TotalSum - K)
# If the subarray with
# sum doesn't exist
if (maxLen == -1):
print (-1,end="")
# Otherwise, prthe
# required maxmimum length
else:
print (N - maxLen,end="")
# Driver Code
if __name__ == '__main__':
arr = [1, 3, 1, 1, 2]
K = 4
N = len(arr)
minRequiredOperation(arr, N, K)
# this code is contributed by mohit kumar 29.C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the length of
// longest subarray having sum K
static int longestSubarray(int []arr,
int N, int K)
{
// Stores the index of
// the prefix sum
Dictionary um = new Dictionary();
int sum = 0, maxLen = 0;
// Traverse the given array
for (int i = 0; i < N; i++) {
// Update sum
sum += arr[i];
// If the subarray starts
// from index 0
if (sum == K)
maxLen = i + 1;
// Add the current prefix sum
// with index if it is not
// present in the map um
if (!um.ContainsKey(sum))
um[sum] = i;
// Check if sum - K is
// present in Map um or not
if (um.ContainsKey(sum - K)) {
// Update the maxLength
if (maxLen < (i - um[sum - K]))
maxLen = i - um[sum - K];
}
}
// Return the required maximum length
return maxLen;
}
// Function to find the minimum removal of
// array elements required to reduce K to 0
static void minRequiredOperation(int []arr,
int N, int K)
{
// Stores the sum of the array
int TotalSum = 0;
// Traverse the array arr[]
for (int i = 0; i < N; i++)
// Update sum of the array
TotalSum += arr[i];
// Find maxLen
int maxLen = longestSubarray(
arr, N, TotalSum - K);
// If the subarray with
// sum doesn't exist
if (maxLen == -1) {
Console.WriteLine(-1);
}
// Otherwise, print the
// required maxmimum length
else
Console.WriteLine(N - maxLen);
}
// Driver Code
public static void Main()
{
int []arr = { 1, 3, 1, 1, 2 };
int K = 4;
int N = arr.Length;
minRequiredOperation(arr, N, K);
}
}
// This code is contributed by SUREDRA_GANGWAR. 输出：
2时间复杂度： O(N)
辅助空间： O(N)