24. Advantage Shuffle

class Solution {
    public int[] advantageCount(int[] A, int[] B) {
        int n = A.length;
        //To stores original indexes of B
        Map<Integer, Queue<Integer>> indexes = new HashMap<>();
        for(int i = 0; i < n; i++){
            indexes.putIfAbsent(B[i], new LinkedList<>());
            indexes.get(B[i]).add(i);
        }
        
        //Sort the array
        Arrays.sort(A);
        Arrays.sort(B);
        
        int i = 0, j = 0;
        int[] result = new int[n];
        Arrays.fill(result, -1);
        
        //Ordering A[i] where possible or adding unused numbers in Queue
        Queue<Integer> unusedNums = new LinkedList<>();
        while(i < n && j < n){
            if(A[i] > B[j]) {
                int ind = indexes.get(B[j]).poll();
                result[ind]=A[i];
                j++;
            }
            else {
                unusedNums.add(A[i]);
            }
            i++;
        }
        
        //Fill the -1's with unusedNums
        for(int in = 0; in < n; in++) {
            if(result[in] == -1){
                result[in] = unusedNums.poll();
            }
        }
        
        return result;
    }
}