The most common interview pattern is: "I need to look up a value I saw earlier in O(1)." A hash map is the answer.

Without a hash map, looking up "have I seen X before?" requires scanning all previous elements: O(n) per lookup → O(n²) total. With a hash map: O(1) average per lookup → O(n) total.

The canonical example — Two Sum

// O(n²) brute force
function twoSum(nums: number[], target: number): number[] {
  for (let i = 0; i < nums.length; i++)
    for (let j = i+1; j < nums.length; j++)
      if (nums[i] + nums[j] === target) return [i, j];
  return [];
}

// O(n) with hash map
function twoSum(nums: number[], target: number): number[] {
  const seen = new Map<number, number>();  // value → index
  for (let i = 0; i < nums.length; i++) {
    const complement = target - nums[i];
    if (seen.has(complement)) return [seen.get(complement)!, i];
    seen.set(nums[i], i);
  }
  return [];
}

Trace the hash version on nums = [3, 7, 1], target = 10:

One pass, each step O(1). Hash maps trade space for time: you pay O(n) space to buy O(n) time instead of O(n²) — the same conscious swap from the Big-O concept.