algorithmfree

Brute-force pair sum

Try every pair to find two numbers that add up to a target. Simple, but O(n²).

time: O(n²)
space: O(1)

// step through it

step 1 / 10

1function twoSumBrute(nums, target) {2  for (let i = 0; i < nums.length; i++) {3    for (let j = i + 1; j < nums.length; j++) {4      if (nums[i] + nums[j] === target) return [i, j];5    }6  }7  return null;8}

target: = 10
checks: = 1

1 + 3 = 4, not 10. Checks so far: 1.

Tip: use the arrow keys to step.

How it works

Two nested loops try every pair: the first number with each one after it, then the second, and so on. For n items that is about n²/2 pairs, so doubling the array roughly quadruples the work. That is O(n²). Compare it with the two-pointers lesson, which solves the same problem on a sorted array in a single pass.