Two Sum - Practice Coding Problems

Two Sum - Problem

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,7,11,15], target = 9

› Output: [0,1]

💡 Note: nums[0] + nums[1] = 2 + 7 = 9, so we return the indices [0,1]

Example 2 — Different Position

$ Input: nums = [3,2,4], target = 6

› Output: [1,2]

💡 Note: nums[1] + nums[2] = 2 + 4 = 6, so we return the indices [1,2]

Example 3 — Duplicate Values

$ Input: nums = [3,3], target = 6

› Output: [0,1]

💡 Note: Both elements are the same: nums[0] + nums[1] = 3 + 3 = 6

Constraints

2 ≤ nums.length ≤ 10⁴
-10⁹ ≤ nums[i] ≤ 10⁹
-10⁹ ≤ target ≤ 10⁹
Only one valid answer exists.

Visualization

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Understanding the Visualization

Input

Array [2,7,11,15] and target 9

Process

Find two numbers that sum to target

Output

Return indices [0,1] of the two numbers

Key Takeaway

🎯 Key Insight: Use a hash map to remember what you've seen, so you can instantly find the complement

Asked in

G Google 127 a Amazon 98 Apple 76 m Microsoft 65

The key insight is to use a hash map to remember previously seen numbers. Instead of checking all pairs (O(n²)), we can find complements in O(1) time. Best approach is hash map with Time: O(n), Space: O(n).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Check All Pairs	O(n²)	O(1)	Try every possible pair of numbers to find the target sum
Hash Map - One Pass	O(n)	O(n)	Use a hash map to remember seen values and instantly find complements

Brute Force - Check All Pairs — Algorithm Steps

Iterate through each element with index i
For each i, iterate through remaining elements with index j
Check if nums[i] + nums[j] equals target
Return [i, j] when found

Visualization

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Step-by-Step Walkthrough

Outer Loop

Start with first element

Inner Loop

Compare with each remaining element

Check Sum

Test if current pair sums to target

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int* solution(int* nums, int numsSize, int target, int* returnSize) {
    for (int i = 0; i < numsSize; i++) {
        for (int j = i + 1; j < numsSize; j++) {
            if (nums[i] + nums[j] == target) {
                int* result = (int*)malloc(2 * sizeof(int));
                result[0] = i;
                result[1] = j;
                *returnSize = 2;
                return result;
            }
        }
    }
    *returnSize = 0;
    return NULL;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[,]");
    while (token != NULL) {
        int num = atoi(token);
        if (num != 0 || token[0] == '0') {
            nums[numsSize++] = num;
        }
        token = strtok(NULL, "[,]");
    }
    
    int target;
    scanf("%d", &target);
    
    int returnSize;
    int* result = solution(nums, numsSize, target, &returnSize);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        if (i > 0) printf(",");
        printf("%d", result[i]);
    }
    printf("]\n");
    
    if (result) free(result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops iterate through n×n combinations in worst case

⚠ Quadratic Growth

Space Complexity

O(1)

Only uses a few variables, no extra data structures

✓ Linear Space

2.1M Views

Very High Frequency

~15 min Avg. Time

15.4K Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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