Two Sum

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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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

✓ Brute Force - Check All Pairs

⏱️ Time: O(n²) Space: O(1)

Use nested loops to check every combination of two different elements. For each element, check if it can pair with any element that comes after it to reach the target sum.

Hash Map - One Pass

⏱️ Time: O(n) Space: O(n)

Iterate through the array once. For each element, check if its complement (target - current) exists in the hash map. If yes, return the indices. If no, store the current element and its index in the hash map.

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 - improved parsing for negative numbers
    int nums[1000];
    int numsSize = 0;
    int i = 0;
    int len = strlen(line);
    
    // Skip opening bracket
    while (i < len && line[i] != '[') i++;
    i++;
    
    while (i < len && line[i] != ']') {
        while (i < len && (line[i] == ' ' || line[i] == ',')) i++;
        if (i < len && line[i] != ']') {
            int sign = 1;
            int num = 0;
            if (line[i] == '-') {
                sign = -1;
                i++;
            }
            while (i < len && line[i] >= '0' && line[i] <= '9') {
                num = num * 10 + (line[i] - '0');
                i++;
            }
            nums[numsSize++] = sign * num;
        }
    }
    
    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

✓ Linear Growth

Space Complexity

O(1)

Only uses a few variables, no extra data structures

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Check All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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