Top Two Pointers Interview Questions & Solutions (2026)

1. Introduction to the Two Pointers Pattern

The Two Pointers technique utilizes two coordinate indices to traverse one or more data structures simultaneously, drastically reducing time or space complexity.

+─────────────────────────────────────────────────────────────────────────────+
|                         TWO POINTERS CONFIGURATIONS                         |
+─────────────────────────────────────────────────────────────────────────────+

 1. Converging (Opposite Direction):
    Left ───> [ 1,  2,  4,  6,  8,  11 ] <─── Right

 2. Fast & Slow (Same Direction):
    Slow ──> [ 1,  1,  2,  3,  3,  4 ]
    Fast ─────────>

2. Core Interview Questions & Verified Solutions

Q1. 3Sum (Converging Pointers Pattern)

Problem Statement: Given an integer array nums, return all triplets [nums[i], nums[j], nums[k]] such that i ≠ j, i ≠ k, j ≠ k and nums[i] + nums[j] + nums[k] == 0. The solution set must not contain duplicate triplets.

python
def threeSum(nums: list[int]) -> list[list[int]]:
    nums.sort()
    result = []
    
    for i in range(len(nums) - 2):
        # Skip duplicate values for first element
        if i > 0 and nums[i] == nums[i - 1]:
            continue
            
        left, right = i + 1, len(nums) - 1
        while left < right:
            total = nums[i] + nums[left] + nums[right]
            
            if total == 0:
                result.append([nums[i], nums[left], nums[right]])
                while left < right and nums[left] == nums[left + 1]:
                    left += 1 # Skip duplicate lefts
                while left < right and nums[right] == nums[right - 1]:
                    right -= 1 # Skip duplicate rights
                left += 1
                right -= 1
            elif total < 0:
                left += 1
            else:
                right -= 1
                
    return result

• Time Complexity: O(N²) | Space Complexity: O(1) auxiliary space.

Q2. Container With Most Water

Problem Statement: Given an integer array height representing vertical lines on the x-axis, find two lines that together with the x-axis form a container containing the most water.

cpp
#include <vector>
#include <algorithm>

int maxArea(const std::vector<int>& height) {
    int left = 0, right = height.size() - 1;
    int maxWater = 0;

    while (left < right) {
        int width = right - left;
        int currentHeight = std::min(height[left], height[right]);
        maxWater = std::max(maxWater, width * currentHeight);

        // Greedily move the shorter line inward
        if (height[left] < height[right]) {
            left++;
        } else {
            right--;
        }
    }
    return maxWater;
}

• Time Complexity: O(N) | Space Complexity: O(1)

Q3. Trapping Rain Water

Problem Statement: Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.

java
public int trap(int[] height) {
    if (height == null || height.length == 0) return 0;

    int left = 0, right = height.length - 1;
    int leftMax = 0, rightMax = 0;
    int totalWater = 0;

    while (left < right) {
        if (height[left] < height[right]) {
            if (height[left] >= leftMax) {
                leftMax = height[left];
            } else {
                totalWater += leftMax - height[left];
            }
            left++;
        } else {
            if (height[right] >= rightMax) {
                rightMax = height[right];
            } else {
                totalWater += rightMax - height[right];
            }
            right--;
        }
    }
    return totalWater;
}

• Time Complexity: O(N) | Space Complexity: O(1)

3. Two Pointers Pattern Summary

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