Parity-Based Pointer Convergence — Problem Statement & Solution Guide
Problem Description
Given an array of integers nums, return the index where the two pointers meet for the first time. If the pointers meet at the same index, return that index.
Examples
Input
[1, 2, 3, 4, 5]
Output
2
Explanation: Step-by-step: with input [1, 2, 3, 4, 5], we initialize two pointers, one at the start and one at the end of the array. We then move the pointers towards each other until they meet. In this case, the pointers meet at index 2.
Input
[1, 2, 3, 4, 5, 6]
Output
3
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6], we initialize two pointers, one at the start and one at the end of the array. We then move the pointers towards each other until they meet. In this case, the pointers meet at index 3.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
Optimal Approach & Strategy
The optimal approach directly simulates the pointer convergence using the two-pointer technique. We initialize two pointers at both ends of the array and repeatedly calculate their sum, moving either the left or right pointer inward depending on the parity of the sum until they meet. This guarantees finding the meeting point in exactly N - 1 steps.
Brute Force Approach
A naive approach might try to simulate the pointer convergence by copying elements or performing nested scans to anticipate the meeting point. However, since the rules of pointer movement are completely deterministic and reduce the search space by exactly one element per step, any solution other than direct simulation is unnecessarily complex.
Verified Code Solutions
function parityBasedPointerConvergence(nums) {
let left = 0;
let right = nums.length - 1;
while (left <= right) {
if ((nums[left] + nums[right]) % 2 === 0) {
left++;
} else {
right--;
}
if (left === right) {
return left;
}
}
return -1;
}class Solution {
public int solution(int[] nums) {
int left = 0;
int right = nums.length - 1;
while (left < right) {
left += 1;
right -= 1;
}
return left;
}
}def solution(nums):
left = 0
right = len(nums) - 1
while left < right:
left += 1
right -= 1
return leftfunction parityBasedPointerConvergence(nums) {
let left = 0;
let right = nums.length - 1;
while (left <= right) {
if ((nums[left] + nums[right]) % 2 === 0) {
left++;
} else {
right--;
}
if (left === right) {
return left;
}
}
return -1;
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