给定一个长度为N的数组arr[] ,任务是找到索引对 (i, j) (基于 0 的索引)的计数,使得子数组{arr[0], … arr[i] 的前缀和}等于子数组{arr[N – 1], …, arr[j]} ( 0 ≤ i, j < N) 的后缀和。
例子:
Input: arr[] = {1, 2, 1, 1}
Output: 3
Explanation:
The 3 possible pairs of indices are as follows:
- Pair {0, 3}: Prefix Sum of subarray {arr[0]} = 1. Suffix Sum of subarray {arr[3]} = 1.
- Pair {2, 1}: Prefix Sum of subarray {arr[0], .. arr[2]} = 4. Suffix Sum of subarray {arr[3], …, arr[1]} = 4.
- Pair {3, 0}: Prefix Sum of subarray {arr[0], .. arr[3]} = 5. Suffix Sum of subarray {arr[3], …, arr[0]} = 5.
Input: arr[] = {1, 0, -1}
Output: 1
做法:思路是用Hashing来解决这个问题。请按照以下步骤解决问题:
- 遍历数组arr[]并计算所有数组索引的前缀和并将它们的频率存储在 HashMap 中。
- 反向遍历数组并继续计算每个数组索引的后缀总和。
- 对于获得的每个后缀和,检查它是否存在于 Map 中。如果发现为真,则将count增加1 。
- 打印获得的计数。
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 count of
// index pairs having equal
// prefix and suffix sums
void countPairs(int* arr, int n)
{
// Maps indices with prefix sums
unordered_map mp1;
int sum = 0;
// Traverse the array
for (int i = 0; i < n; i++) {
// Update prefix sum
sum += arr[i];
// Update frequency in Map
mp1[sum] += 1;
}
sum = 0;
int ans = 0;
// Traverse the array in reverse
for (int i = n - 1; i >= 0; i--) {
// Update suffix sum
sum += arr[i];
// Check if any prefix sum of
// equal value exists or not
if (mp1.find(sum) != mp1.end()) {
ans += mp1[sum];
}
}
// Print the obtained count
cout << ans;
}
// Driver code
int main()
{
// Given array
int arr[] = { 1, 2, 1, 1 };
// Given size
int n = sizeof(arr)
/ sizeof(arr[0]);
// Function Call
countPairs(arr, n);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG
{
// Function to find the count of
// index pairs having equal
// prefix and suffix sums
static void countPairs(int []arr, int n)
{
// Maps indices with prefix sums
HashMap mp1 = new HashMap();
int sum = 0;
// Traverse the array
for (int i = 0; i < n; i++)
{
// Update prefix sum
sum += arr[i];
// Update frequency in Map
if(mp1.containsKey(sum)){
mp1.put(sum, mp1.get(sum)+1);
}
else{
mp1.put(sum, 1);
}
}
sum = 0;
int ans = 0;
// Traverse the array in reverse
for (int i = n - 1; i >= 0; i--)
{
// Update suffix sum
sum += arr[i];
// Check if any prefix sum of
// equal value exists or not
if (mp1.containsKey(sum))
{
ans += mp1.get(sum);
}
}
// Print the obtained count
System.out.print(ans);
}
// Driver code
public static void main(String[] args)
{
// Given array
int arr[] = { 1, 2, 1, 1 };
// Given size
int n = arr.length;
// Function Call
countPairs(arr, n);
}
}
// This code is contributed by Amit Katiyar
Python3
# Python3 program for the above approach
# Function to find the count of
# index pairs having equal
# prefix and suffix sums
def countPairs(arr, n):
# Maps indices with prefix sums
mp1 = {}
sum = 0
# Traverse the array
for i in range(n):
# Update prefix sum
sum += arr[i]
# Update frequency in Map
mp1[sum] = mp1.get(sum, 0) + 1
sum = 0
ans = 0
# Traverse the array in reverse
for i in range(n - 1, -1, -1):
# Update suffix sum
sum += arr[i]
# Check if any prefix sum of
# equal value exists or not
if (sum in mp1):
ans += mp1[sum]
# Print the obtained count
print (ans)
# Driver code
if __name__ == '__main__':
# Given array
arr = [ 1, 2, 1, 1 ]
# Given size
n = len(arr)
# Function Call
countPairs(arr, n)
# This code is contributed by mohit kumar 29
C#
// C# code for above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find the count of
// index pairs having equal
// prefix and suffix sums
static void countPairs(int[] arr, int n)
{
// Maps indices with prefix sums
Dictionary mp1 = new Dictionary();
int sum = 0;
// Traverse the array
for (int i = 0; i < n; i++)
{
// Update prefix sum
sum += arr[i];
// Update frequency in Map
if(mp1.ContainsKey(sum))
{
mp1.Add(sum, mp1[sum] + 1);
}
else{
mp1.Add(sum, 1);
}
}
sum = 0;
int ans = 0;
// Traverse the array in reverse
for (int i = n - 1; i >= 0; i--)
{
// Update suffix sum
sum += arr[i];
// Check if any prefix sum of
// equal value exists or not
if (mp1.ContainsKey(sum))
{
ans += mp1[sum];
}
}
// Print the obtained count
Console.Write(ans);
}
// Driver code
static public void Main ()
{
// Given array
int[] arr = { 1, 2, 1, 1 };
// Given size
int n = arr.Length;
// Function Call
countPairs(arr, n);
}
}
// This code is contributed by offbeat
输出:
3
时间复杂度: O(N)
辅助空间: O(N)
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