查找数组元素，它等于至少大小为 2 的任何子数组之和
原文：https://www.geeksforgeeks.org/find-array-elements-equal-to-sum-of-any-subarray-of-at-least-size-2/
给定一个数组arr[]，任务是从该数组中查找等于大小大于 1 的任何子数组之和的元素。
示例：
输入：
arr [] = {1, 2, 3, 4, 5, 6}输出：
3, 5, 6说明：
元素 3、5、6 分别等于子数组
{1, 2}，{2, 3}和{1, 2, 3}的总和。输入：
arr [] = {5, 6, 10, 1, 3, 4, 8, 16}输出：
4, 8, 16解释：
元素 4、8、16 分别等于子数组
{1, 3}, {1, 3, 4}, {1, 3, 4, 8}的总和。
方法：想法是使用前缀和数组解决给定的问题。 创建一个prefix[]数组，该数组为每个索引存储其所有先前元素的前缀和。 将所有子数组的总和存储在映射中，并搜索映射中是否存在任何数组元素。
以下是上述方法的实现：
C++
// C++ implementation to find array
// elements equal to the sum
// of any subarray of at least size 2
#include <bits/stdc++.h>
using namespace std;
// Function to find all elements
void findElements(int* arr, int n)
{
if (n == 1)
return;
int pre[n];
// Create a prefix array
pre[0] = arr[0];
for (int i = 1; i < n; i++)
pre[i] = arr[i] + pre[i - 1];
unordered_map<int, bool> mp;
// Mark the sum of all sub arrays as true
for (int i = 1; i < n - 1; i++) {
mp[pre[i]] = true;
for (int j = i + 1; j < n; j++) {
mp[pre[j] - pre[i - 1]] = true;
}
}
// Check all elements
// which are marked
// true in the map
for (int i = 0; i < n; i++) {
if (mp[arr[i]]) {
cout << arr[i] << " ";
}
}
cout << endl;
}
// Driver Code
int main()
{
int arr1[] = { 1, 2, 3, 4, 5, 6 };
int n1 = sizeof(arr1) / sizeof(arr1[0]);
findElements(arr1, n1);
return 0;
}
Java
// Java implementation to find array
// elements equal to the sum of any
// subarray of at least size 2
import java.util.*;
class GFG{
// Function to find all elements
static void findElements(int[] arr, int n)
{
if (n == 1)
return;
int pre[] = new int[n];
// Create a prefix array
pre[0] = arr[0];
for(int i = 1; i < n; i++)
pre[i] = arr[i] + pre[i - 1];
HashMap<Integer, Boolean> mp = new HashMap<>();
// Mark the sum of all sub arrays as true
for(int i = 1; i < n - 1; i++)
{
mp.put(pre[i], true);
for(int j = i + 1; j < n; j++)
{
mp.put(pre[j] - pre[i - 1], true);
}
}
// Check all elements
// which are marked
// true in the map
for(int i = 0; i < n; i++)
{
if (mp.get(arr[i]) != null)
{
System.out.print(arr[i] + " ");
}
}
System.out.println();
}
// Driver Code
public static void main(String[] args)
{
int arr1[] = { 1, 2, 3, 4, 5, 6 };
int n1 = arr1.length;
findElements(arr1, n1);
}
}
// This code is contributed by jrishabh99
Python3
# Python3 implementation to find array
# elements equal to the sum of any
# subarray of at least size 2
# Function to find all elements
def findElements(arr, n):
if (n == 1):
return;
pre = [0] * n;
# Create a prefix array
pre[0] = arr[0];
for i in range(1, n):
pre[i] = arr[i] + pre[i - 1];
mp = {};
# Mark the sum of all sub arrays as true
for i in range(1, n - 1):
mp[pre[i]] = True;
for j in range(i + 1, n):
mp[pre[j] - pre[i - 1]] = True;
# Check all elements which
# are marked true in the map
for i in range(n):
if arr[i] in mp:
print(arr[i], end = " ");
else:
pass
print()
# Driver Code
if __name__ == "__main__":
arr1 = [ 1, 2, 3, 4, 5, 6 ];
n1 = len(arr1);
findElements(arr1, n1);
# This code is contributed by AnkitRai01
C
// C# implementation to find array
// elements equal to the sum of any
// subarray of at least size 2
using System;
using System.Collections.Generic;
class GFG{
// Function to find all elements
static void findElements(int[] arr,
int n)
{
if (n == 1)
return;
int []pre = new int[n];
// Create a prefix array
pre[0] = arr[0];
for(int i = 1; i < n; i++)
pre[i] = arr[i] + pre[i - 1];
Dictionary<int,
Boolean> mp = new Dictionary<int,
Boolean>();
// Mark the sum of all sub arrays as true
for(int i = 1; i < n - 1; i++)
{
if(!mp.ContainsKey(pre[i]))
mp.Add(pre[i], true);
for(int j = i + 1; j < n; j++)
{
if(!mp.ContainsKey(pre[j] - pre[i - 1]))
mp.Add(pre[j] - pre[i - 1], true);
}
}
// Check all elements
// which are marked
// true in the map
for(int i = 0; i < n; i++)
{
if (mp.ContainsKey(arr[i]))
{
Console.Write(arr[i] + " ");
}
}
Console.WriteLine();
}
// Driver Code
public static void Main(String[] args)
{
int []arr1 = {1, 2, 3, 4, 5, 6};
int n1 = arr1.Length;
findElements(arr1, n1);
}
}
// This code is contributed by gauravrajput1
输出：
3 5 6
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