Two Pointer Technique on Arrays in DSA C

Mental Model

Use two markers moving through the array to find pairs or conditions without checking every pair.

Analogy: Imagine two friends starting at opposite ends of a hallway, walking towards each other to find a matching pair of shoes quickly.

Array: [1] -> [2] -> [3] -> [4] -> [5] -> null
Pointers: ↑left                             ↑right

Dry Run Walkthrough

Input: array: [1, 2, 3, 4, 5], find if any two numbers sum to 7

Goal: Find two numbers in the array that add up to 7 using two pointers

Step 1: Set left pointer at index 0 (value 1), right pointer at index 4 (value 5)

[1]↑left -> [2] -> [3] -> [4] -> [5]↑right -> null

Why: Start with pointers at both ends to check sums efficiently

Step 2: Calculate sum = 1 + 5 = 6, which is less than 7, move left pointer right

[1] -> [2]↑left -> [3] -> [4] -> [5]↑right -> null

Why: Sum too small, increase sum by moving left pointer forward

Step 3: Calculate sum = 2 + 5 = 7, found target sum, stop

[1] -> [2]↑left -> [3] -> [4] -> [5]↑right -> null

Why: Sum matches target, solution found

Result:

[1] -> [2] -> [3] -> [4] -> [5] -> null
Pair found: 2 + 5 = 7

Annotated Code

DSA C

#include <stdio.h>
#include <stdbool.h>

bool two_pointer_sum(int arr[], int size, int target) {
    int left = 0;
    int right = size - 1;
    while (left < right) {
        int sum = arr[left] + arr[right];
        if (sum == target) {
            printf("Pair found: %d + %d = %d\n", arr[left], arr[right], target);
            return true;
        } else if (sum < target) {
            left++; // increase sum by moving left pointer right
        } else {
            right--; // decrease sum by moving right pointer left
        }
    }
    printf("No pair found that sums to %d\n", target);
    return false;
}

int main() {
    int arr[] = {1, 2, 3, 4, 5};
    int size = sizeof(arr) / sizeof(arr[0]);
    int target = 7;
    two_pointer_sum(arr, size, target);
    return 0;
}

while (left < right) {

loop until pointers meet or cross

int sum = arr[left] + arr[right];

calculate sum of values at pointers

if (sum == target) {

check if sum matches target

left++; // increase sum by moving left pointer right

move left pointer right to increase sum when sum too small

right--; // decrease sum by moving right pointer left

move right pointer left to decrease sum when sum too large

Output

Pair found: 2 + 5 = 7

Complexity Analysis

Time: O(n) because each pointer moves at most n steps through the array once

Space: O(1) because only two pointers and a few variables are used

vs Alternative: Compared to checking all pairs (O(n^2)), this is much faster by avoiding repeated checks

Edge Cases

empty array

function returns no pair found immediately

DSA C

while (left < right) {

array with one element

function returns no pair found because pointers cannot cross

DSA C

while (left < right) {

no two numbers sum to target

function prints no pair found and returns false

DSA C

printf("No pair found that sums to %d\n", target);

When to Use This Pattern

When you need to find pairs or conditions involving two elements in a sorted array, use two pointers moving inward to find answers efficiently.

Common Mistakes

Mistake: Not moving the correct pointer based on sum comparison
Fix: If sum is less than target, move left pointer right; if sum is greater, move right pointer left

Summary

Find pairs in a sorted array that meet a condition using two pointers.

Use when searching pairs or conditions in sorted arrays to reduce time.

Move pointers inward based on sum comparison to avoid checking all pairs.