LeetCode 3343: Count Number of Balanced Permutations Solution

Master LeetCode problem 3343 (Count Number of Balanced Permutations), 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.

3343. Count Number of Balanced Permutations

Hard

Math String Dynamic Programming Combinatorics

Problem Explanation

Explanation

To solve this problem, we need to find all distinct permutations of the given string num and count how many of them are balanced. We can achieve this by using backtracking to generate all permutations and then checking if each permutation is balanced.

Here is the high-level approach:

Implement a helper function to check if a given permutation is balanced.
Use backtracking to generate all permutations of the input string.
For each permutation, check if it is balanced and increment the count if it is.
Return the count modulo 10^9 + 7 as the final answer.

Time complexity: O(n!) where n is the length of the input string num. Space complexity: O(n) for storing the permutations and additional space for recursion stack.

Solution Code

class Solution {
    private static final int MOD = 1000000007;
    private int count;

    public int countNumberPermutations(String num) {
        count = 0;
        char[] arr = num.toCharArray();
        boolean[] used = new boolean[arr.length];
        generatePermutations(arr, used, new StringBuilder());
        return count;
    }

    private void generatePermutations(char[] arr, boolean[] used, StringBuilder current) {
        if (current.length() == arr.length) {
            if (isBalanced(current.toString())) {
                count = (count + 1) % MOD;
            }
            return;
        }

        for (int i = 0; i < arr.length; i++) {
            if (!used[i]) {
                used[i] = true;
                current.append(arr[i]);
                generatePermutations(arr, used, current);
                current.deleteCharAt(current.length() - 1);
                used[i] = false;
            }
        }
    }

    private boolean isBalanced(String str) {
        int sumOdd = 0;
        int sumEven = 0;
        for (int i = 0; i < str.length(); i++) {
            if (i % 2 == 0) {
                sumEven += str.charAt(i) - '0';
            } else {
                sumOdd += str.charAt(i) - '0';
            }
        }
        return sumOdd == sumEven;
    }
}

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