Complement Lookup

The canonical example is Two Sum on an unsorted array — no sorting, so the opposite-ends trick from the two-pointers pattern doesn't apply, and a hash map fills in instead.

int* twoSum(int *nums, int n, int target) {
      HashMap seen = {0};   /* value -> index */
      for (int i = 0; i < n; i++) {
          int complement = target - nums[i];
          if (seen.contains(complement)) {
              return (int[]){ seen[complement], i };
          }
          seen[nums[i]] = i;
      }
      return NULL;
  }

The order matters: check for the complement before storing the current value, not after — otherwise a target that needs the same element twice would match against itself. First pass at this used range(len(nums) - 1), which quietly skips the last element; the fix is just range(len(nums)), since there's no reason the last element can't be the one that completes a pair.