All Unique Permutations of an array

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Problem Overview

Generate all unique permutations of array with duplicates.

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Intuition

Generate all unique permutations of array with duplicates. Sort + backtrack with skip-duplicates logic.

Algorithm

1Sort array. Backtrack: for each position, skip if arr[i]==arr[i-1] and i-1 not used. Swap and recurse.

Common Pitfalls

• Same as LC 47. Must sort first. Skip duplicate choices at same recursion level to avoid repeat permutations.

All Unique Permutations of an array.java

Java

// Approach: Sort array, then backtrack with a 'used' boolean array.
// Skip duplicate elements at the same recursion depth to avoid duplicate permutations.
// Time: O(n! * n) Space: O(n)
import java.util.*;

class Solution {
    public static ArrayList<ArrayList<Integer>> uniquePerms(int[] arr) {
        ArrayList<ArrayList<Integer>> res = new ArrayList<>();
        Arrays.sort(arr);
        do {
            ArrayList<Integer> perm = new ArrayList<>(arr.length);
            
            for (int x : arr)
                perm.add(x);
            res.add(perm);

        } while (nextPermutation(arr));

        return res;
    }

    private static boolean nextPermutation(int[] a) {
        int n = a.length;
        int i = n - 2;
        while (i >= 0 && a[i] >= a[i + 1])
            i--;

        if (i < 0)
            return false;

        int j = n - 1;
        while (a[j] <= a[i])
            j--;

        swap(a, i, j);
        reverse(a, i + 1, n - 1);

        return true;
    }

    private static void swap(int[] a, int i, int j) {
        int t = a[i];
        a[i] = a[j];
        a[j] = t;
    }

    private static void reverse(int[] a, int l, int r) {
        while (l < r)
            swap(a, l++, r--);
    }
};

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