查找前缀总和大于后缀总和的 Array 的所有索引
给定一个包含N个整数的数组arr[] ，任务是除以找到所有索引，使得前缀总和（即范围[0, i)中的元素总和）大于后缀总和（范围[i, N-1] )
例子：
Input: arr = [10, -3, 4, 6]
Output: [1, 3]
Explanation: Consider index 1. Prefix sum = 10, suffix sum = 7, i.e. (10 > 7)
Consider index 3. Prefix sum = 11, suffix sum = 6, i.e(11 > 6)
Input: arr = [-2, -3, -4, 10]
Output: []
Explanation: There is no index such that prefix sum is greater than suffix sum.
方法：基本思想是考虑每个索引并为此计算前缀和后缀总和。如果前缀总和大于后缀总和，则将此索引插入答案。
下面是上述方法的实现。
C++
// C++ code to implement the approach
#include
using namespace std;
class Solution
{
public:
// Function to find the valid indices
vector solve(vector& arr)
{
vector ans;
// Total_size of the array
int size = arr.size();
int left_sum = 0;
int total_sum = 0;
for (auto dt : arr)
total_sum += dt;
// Upto second last index
for (int idx = 0; idx < size - 1; ++idx) {
// Add current element to
// left_sum
left_sum += arr[idx];
// Calculate right_sum
int right_sum = total_sum - left_sum;
// Check condition
if (left_sum > right_sum)
ans.push_back(idx + 1);
}
return (ans);
}
};
// Driver code
int main()
{
Solution obj;
vector arr = { 10, -3, 4, 6 };
vector ans = obj.solve(arr);
for (auto x : ans)
cout << x << " ";
return 0;
}
// This code is contributed by rakeshsahnis Java
// Java code to implement the approach
import java.util.*;
class GFG {
// Function to find the valid indices
static List solve(int arr[])
{
List ans = new ArrayList();
// Total_size of the array
int size = arr.length;
int left_sum = 0;
int total_sum = 0;
for(int i = 0; i < size; i++){
total_sum += arr[i];
}
// Upto second last index
for (int idx = 0; idx < size - 1; ++idx) {
// Add current element to
// left_sum
left_sum += arr[idx];
// Calculate right_sum
int right_sum = total_sum - left_sum;
// Check condition
if (left_sum > right_sum)
ans.add(idx + 1);
}
return ans;
}
// Driver code
public static void main (String[] args) {
int arr[] = { 10, -3, 4, 6 };
List ans = solve(arr);
for (Integer x : ans)
System.out.print(x + " ");
}
}
// This code is contributed by hrithikgarg03188. Python3
# Python code to implement the above approach
class Solution:
# Function to find the indices
def solve(self, arr):
ans = []
# Total_size of the array
size = len(arr)
# Upto second last index
for idx in range(size-1):
left_sum = 0
right_sum = 0
# Calculate left sum
for left_idx in range(idx + 1):
left_sum += arr[left_idx]
# Calculate right sum
for right_idx in range(idx + 1,
size, 1):
right_sum += arr[right_idx]
# check condition
if (left_sum > right_sum):
ans.append(idx + 1)
return(ans)
# Driver code
if __name__ == '__main__':
obj = Solution()
arr = [10, -3, 4, 6]
ans = obj.solve(arr)
for x in ans:
print (x, end = " ")C#
// C# code to implement the approach
using System;
using System.Collections;
class GFG {
// Function to find the valid indices
static ArrayList solve(int[] arr)
{
ArrayList ans = new ArrayList();
// Total_size of the array
int size = arr.Length;
int left_sum = 0;
int total_sum = 0;
for (int i = 0; i < size; i++) {
total_sum += arr[i];
}
// Upto second last index
for (int idx = 0; idx < size - 1; ++idx) {
// Add current element to
// left_sum
left_sum += arr[idx];
// Calculate right_sum
int right_sum = total_sum - left_sum;
// Check condition
if (left_sum > right_sum)
ans.Add(idx + 1);
}
return ans;
}
// Driver code
public static void Main()
{
int[] arr = { 10, -3, 4, 6 };
ArrayList ans = solve(arr);
foreach(int x in ans) Console.Write(x + " ");
}
}
// This code is contributed by Samim Hossain Mondal.Javascript
Python3
# Python code to implement the approach
class Solution:
# Function to find the valid indices
def solve(self, arr):
ans = []
# Total_size of the array
size = len(arr)
left_sum = 0
total_sum = sum(arr)
# Upto second last index
for idx in range(size-1):
# Add current element to
# left_sum
left_sum += arr[idx]
# Calculate right_sum
right_sum = total_sum - left_sum
# Check condition
if (left_sum > right_sum):
ans.append(idx + 1)
return(ans)
# Driver code
if __name__ == '__main__':
obj = Solution()
arr = [10, -3, 4, 6]
ans = obj.solve(arr)
for x in ans:
print (x, end = " ")输出
1 3 时间复杂度： O(N 2 )
辅助空间： O(1)
Efficient Approach ：上述方法可以通过基于以下思想计算数组的总和来进一步优化：
Sum of array = prefix sum at index i + suffix sum at index i. So find the array sum. Then iterate array and keep on calculating the prefix sum. If the condition of the problem is satisfied by prefix and suffix sum then that index will be part of answer.
请按照以下步骤解决问题：
- 迭代数组并求出数组的总和（比如总和）。
- 初始化变量（比如 temp）以存储前缀总和。
- 从i = 1 迭代到 N-1 ：
- 将arr[i-1]添加到前缀 sum。
- 使用上述观察计算后缀和。
- 如果前缀和后缀总和满足问题的条件，我将是答案之一。
- 返回存储有效索引的数组。
下面是上述方法的实现。
Python3
# Python code to implement the approach
class Solution:
# Function to find the valid indices
def solve(self, arr):
ans = []
# Total_size of the array
size = len(arr)
left_sum = 0
total_sum = sum(arr)
# Upto second last index
for idx in range(size-1):
# Add current element to
# left_sum
left_sum += arr[idx]
# Calculate right_sum
right_sum = total_sum - left_sum
# Check condition
if (left_sum > right_sum):
ans.append(idx + 1)
return(ans)
# Driver code
if __name__ == '__main__':
obj = Solution()
arr = [10, -3, 4, 6]
ans = obj.solve(arr)
for x in ans:
print (x, end = " ")
输出
1 3 时间复杂度： O(N)
辅助空间： O(1)