How can you find two indices whose values sum to a target in one pass?

Keep a map from previously seen values to their indices. For each element, check whether its complement, target minus current value, has already appeared. Check before inserting so the same element cannot be used twice. This naturally handles duplicate values at different positions, such as two 3s for target 6. Expected time is O(n), with O(n) extra space under ordinary hash-map assumptions. State what to return when no pair exists and whether one pair or all pairs are required.