在这个问题中，我们给出了一个大小为N的数组arr[]。我们的任务是在数组中可能的移动后找到左指针的索引。

我们有两个指针，一个是左指针，另一个是右指针。

左指针从索引0开始，值递增。

右指针从索引(n-1)开始，值递减。

如果左指针的和小于右指针的和，则指针的值增加，否则指针的值减少。并且和的值会更新。

让我们通过一个例子来理解这个问题，

Input : arr[] = {5, 6, 3, 7, 9, 4}
Output : 2

Explanation −

Explanation −

leftPointer = 0 -> sum = 5, rightPointer = 5 -> sum = 4. Move rightPointer
leftPointer = 0 -> sum = 5, rightPointer = 4 -> sum = 13. Move leftPointer
leftPointer = 1 -> sum = 11, rightPointer = 4 -> sum = 13. Move leftPointer
leftPointer = 2 -> sum = 14, rightPointer = 4 -> sum = 13. Move rightPointer
leftPointer = 2 -> sum = 14, rightPointer = 3 -> sum = 20. Move rightPointer
Position of the left pointer is 2.

解决方案

通过根据和的大小移动左指针和右指针来解决问题。然后检查左指针是否比右指针大1。

示例

程序示例，说明我们解决方案的工作原理

#include <iostream>
using namespace std;
int findIndexLeftPointer(int arr[], int n) {
   if(n == 1)
      return 0;
   int leftPointer = 0,rightPointer = n-1,leftPointerSum = arr[0], rightPointerSum = arr[n-1];
   while (rightPointer > leftPointer + 1) {
      if (leftPointerSum < rightPointerSum) {
         leftPointer++;
         leftPointerSum += arr[leftPointer];
      }
      else if (leftPointerSum > rightPointerSum) {
         rightPointer--;
         rightPointerSum += arr[rightPointer];
      }
      else {
         break;
      }
   }
   return leftPointer;
}
int main() {
   int arr[] = { 5, 6, 3, 7, 9, 4 };
   int n = sizeof(arr) / sizeof(arr[0]);
   cout<<"The index of left pointer after moving is "<<findIndexLeftPointer(arr, n);
   return 0;
}

输出

The index of left pointer after moving is 2