import java.util.*;

class Solution {
		
	/**
	*	@param int[] nums An array of non-negative integers
	*	@param int sum The value the pair of elements must sum to
	*
	*	@return The distinct pair of VALUES in the array whose corresponding
	*			values sum to the value of 'sum', or [-1, -1] if not found.
	**/
    public static int[] twoSumValues(int[] nums, int sum) {
		// Create a map of num elements to indices
		Map numToIndex = new HashMap();
		
		// Fill the map
		for (int i = 0; i < nums.length; i++) {
			// It doesn't matter if indices are overwritten
			// because the final value will map to the largest index
			numToIndex.put(nums[i], i);
		}
		
		// Iterate through values from smallest index to largest
		// This ensures that overwritten values still match to later indices
		// (e.g., if sum is 4 and there is a 2 at both indices 0 and 1, map only contains 2 mapped to 1)
		for (int i = 0; i < nums.length - 1; i++) {
			
			// See if a match exists in map
			Integer index = numToIndex.get(sum - nums[i]);
			
			// If match exists and the second element's index is later than the first
			if (index != null && index > i) {
				// Return the pair of distinct indices
				return new int[]{nums[i], nums[index]};
			}
		}
		
		// No solution found
		return new int[]{-1, -1};
    }
	
	/**
	*	@param int[] nums An array of non-negative integers
	*	@param int sum The value 
	*
	*	@return The distinct pair of INDICES in the array whose corresponding
	*			values sum to the value of 'sum', or [-1, -1] if not found.
	**/
    public static int[] twoSumIndices(int[] nums, int sum) {
		// Create a map of num elements to indices
		Map numToIndex = new HashMap();
		
		// Fill the map
		for (int i = 0; i < nums.length; i++) {
			// It doesn't matter if indices are overwritten
			// because the final value will map to the largest index
			numToIndex.put(nums[i], i);
		}
		
		// Iterate through values from smallest index to largest
		// This ensures that overwritten values still match to later indices
		// (e.g., if sum is 4 and there is a 2 at both indices 0 and 1, map only contains 2 mapped to 1)
		for (int i = 0; i < nums.length - 1; i++) {
			
			// See if a match exists in map
			Integer index = numToIndex.get(sum - nums[i]);
			
			// If match exists and the second element's index is later than the first
			if (index != null && index > i) {
				// Return the pair of distinct indices
				return new int[]{i, index};
			}
		}
		
		// No solution found
		return new int[]{-1, -1};
    }
}