LeetCode 1982: Find Array Given Subset Sums Solution

Master LeetCode problem 1982 (Find Array Given Subset Sums), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

1982. Find Array Given Subset Sums

Hard

Array Divide and Conquer

Problem Explanation

Explanation

To solve this problem, we can observe that each element in the unknown array contributes to a certain subset sum. By iterating through the possible values for each element, we can reconstruct the unknown array that generates the given subset sums. We can use backtracking to explore all possible values for each element until we find a valid solution.

Algorithm:

Initialize an array ans of size n to store the unknown array.
Implement a recursive backtracking function that takes the current index idx and the remaining subset sums.
At each recursive call, iterate through all possible values for the current element ans[idx]. For each value, check if it can contribute to any of the remaining subset sums. If it does, recursively call the function with idx+1 and the updated subset sums.
If a valid solution is found (all subset sums are accounted for), return true to stop further exploration.
If no valid solution is found, backtrack by resetting the current element and continue exploring other possibilities.
Call the backtracking function with initial parameters and return the found unknown array.

Time Complexity:

The time complexity of this algorithm is O(2^n) where n is the length of the unknown array. This is because in the worst case, we explore all possible values for each element in the array.

Space Complexity:

The space complexity of this algorithm is O(n) where n is the length of the unknown array. This is due to the space used by the recursive stack.

Solution Code

class Solution {
    public int[] recoverArray(int n, int[] sums) {
        int[] ans = new int[n];
        Arrays.sort(sums);
        backtrack(ans, 0, sums);
        return ans;
    }
    
    private boolean backtrack(int[] ans, int idx, int[] sums) {
        if (idx == ans.length) {
            return true;
        }
        
        for (int i = 0; i < sums.length; i++) {
            if (sums[i] != Integer.MAX_VALUE) {
                int val = sums[i];
                ans[idx] = val;
                sums[i] = Integer.MAX_VALUE;
                
                List<Integer> remaining = new ArrayList<>();
                for (int j = i + 1; j < sums.length; j++) {
                    remaining.add(sums[j] - val);
                }
                
                if (backtrack(ans, idx + 1, remaining.stream().mapToInt(Integer::intValue).toArray())) {
                    return true;
                }
                
                sums[i] = val;
            }
        }
        
        return false;
    }
}

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