Count of Safe XOR Permutations — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Generate all subsets or permutations
O(N! * N)O(N)Problem Description
You are given an integer array nums, which may contain duplicate elements, and an integer array banned. A permutation of nums is considered "safe" if the bitwise XOR sum of any of its prefixes is not present in the banned array. Return the total number of unique safe permutations of nums. Since the array nums can contain duplicate values, you must only count distinct permutations.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Count of Safe XOR Permutations"
WHY DOES IT MATTER?
Backtracking with pruning is essential for problems involving permutations, combinations, or subsets with constraints. It allows exploring the solution space efficiently by abandoning paths that cannot lead to a valid solution.
OPTIMIZATION CHALLENGE
The key optimization is early pruning based on the prefix XOR condition and efficient duplicate handling to avoid redundant permutations.
REAL-WORLD CONNECTION
This pattern is analogous to constraint satisfaction problems in scheduling or routing, where partial solutions are evaluated to avoid generating infeasible schedules or routes.
Always sort the array before backtracking to facilitate duplicate skipping. Use a boolean array or set to track used elements in the current permutation path.
COMPLEXITY AT A GLANCE
O(N! * N)O(N)Core Theory — Why This Approach?
The problem requires counting distinct permutations of an array with duplicates such that no prefix XOR sum matches a banned value. This is a classic combinatorial enumeration problem constrained by a stateful condition (prefix XOR). The naive approach of generating all permutations and filtering them is infeasible for large N due to the factorial growth O(N!) of the permutation space. Even with duplicate handling, the search space remains exponential, necessitating a backtracking strategy that prunes invalid branches early.
Interview Questions on This Problem
Q1How do you handle duplicate elements in the input array to ensure you only count unique permutations?
Sort the array first. During backtracking, when iterating through elements to pick the next one, skip any element that is the same as the previous one if the previous one was not used in the current recursion level. This ensures that identical values are not swapped to create duplicate permutations.
Q2Why is it important to check the prefix XOR condition at each step rather than at the end of the permutation?
Checking at each step allows for early pruning. If a prefix XOR is banned, any permutation extending that prefix is also invalid. This significantly reduces the search space by avoiding the generation of invalid subtrees in the recursion tree.
Q3What data structure would you use to efficiently check if a prefix XOR is banned?
A HashSet (or unordered_set in C++) containing the banned values. This allows for O(1) average time complexity lookups to determine if the current prefix XOR is in the banned set.
Examples
Input
[1, 2, 3], [0, 3]
Output
0
Explanation: Step-by-step: We have three elements [1, 2, 3] and two banned numbers [0, 3]. The XOR sum of any prefix of the permutation [1, 2, 3] will always be present in the banned array {0, 3}. Therefore, the total number of unique safe permutations of nums is 0.
Input
[1], []
Output
1
Explanation: Step-by-step: We have one element [1] and no banned numbers. The permutation [1] is safe. Therefore, the total number of unique safe permutations of nums is 1.
Constraints
- 1 <= nums.length <= 11
- 0 <= nums[i] <= 10^6
- 1 <= banned.length <= 10^5
- 0 <= banned[i] <= 10^6
Optimal Approach & Strategy
Use backtracking to build permutations incrementally. Sort the array and skip duplicates. Maintain a running XOR sum and prune branches where the current prefix XOR is in the banned set. This avoids generating invalid permutations and handles duplicates efficiently.
Brute Force Approach
Generate all possible permutations of the array, including duplicates. For each permutation, calculate the prefix XOR sums and check if any are in the banned array. Count the valid permutations.
Code Solutions
function countSafeXorPermutations(nums, banned) {
const bannedSet = new Set(banned);
const counts = new Map();
for (const n of nums) counts.set(n, (counts.get(n) || 0) + 1);
const uniqueNums = [...new Set(nums)].sort((a, b) => a - b);
function backtrack(currXor, remaining) {
if (remaining === 0) return 1;
let count = 0;
for (const num of uniqueNums) {
if (counts.get(num) > 0) {
const nextXor = currXor ^ num;
if (!bannedSet.has(nextXor)) {
counts.set(num, counts.get(num) - 1);
count += backtrack(nextXor, remaining - 1);
counts.set(num, counts.get(num) + 1);
}
}
}
return count;
}
let safeCount = 0;
for (const num of uniqueNums) {
const countsCopy = new Map(counts);
countsCopy.set(num, countsCopy.get(num) - 1);
safeCount += backtrack(0, uniqueNums.length - 1);
countsCopy.set(num, countsCopy.get(num) + 1);
}
return safeCount;
}#include <vector>
#include <unordered_set>
#include <algorithm>
class Solution {
private:
int n;
std::vector<int> memo;
std::unordered_set<int> banned_set;
std::vector<int> sorted_nums;
int dfs(int mask, int curr_xor) {
if (mask == (1 << n) - 1) {
return 1;
}
if (memo[mask] != -1) {
return memo[mask];
}
int ans = 0;
for (int i = 0; i < n; ++i) {
if ((mask & (1 << i)) == 0) {
// To avoid duplicate permutations, we only use a duplicate element
// if its preceding identical element in the sorted array has already been used.
if (i > 0 && sorted_nums[i] == sorted_nums[i - 1] && (mask & (1 << (i - 1))) == 0) {
continue;
}
int next_xor = curr_xor ^ sorted_nums[i];
if (banned_set.find(next_xor) == banned_set.end()) {
ans += dfs(mask | (1 << i), next_xor);
}
}
}
return memo[mask] = ans;
}
public:
int countSafeXorPermutations(std::vector<int>& nums, std::vector<int>& banned) {
n = nums.size();
sorted_nums = nums;
std::sort(sorted_nums.begin(), sorted_nums.end());
banned_set = std::unordered_set<int>(banned.begin(), banned.end());
memo.assign(1 << n, -1);
return dfs(0, 0);
}
};class Solution {
public int countSafePermutations(int[] nums, int[] banned) {
Set<Integer> xor_set = new HashSet<>();
for (int num : banned) {
xor_set.add(num);
}
xor_set.add(0);
int n = nums.length;
int res = 0;
backtrack(0, 0);
return res;
}
private void backtrack(int idx, int curr_xor) {
if (idx == n) {
res += 1;
return;
}
for (int i = 0; i < n; i++) {
if (!xor_set.contains(nums[i]) && !xor_set.contains(curr_xor ^ nums[i])) {
backtrack(idx + 1, curr_xor ^ nums[i]);
}
}
}
}def countSafePermutations(nums, banned):
xor_set = set()
for num in banned:
xor_set.add(num)
n = len(nums)
xor_set.add(0)
res = 0
def backtrack(idx, curr_xor):
if idx == n:
res += 1
return
for i in range(n):
if i not in xor_set and (curr_xor ^ nums[i]) not in xor_set:
backtrack(idx + 1, curr_xor ^ nums[i])
backtrack(0, 0)
return resfunction countSafeXorPermutations(nums, banned) {
const bannedSet = new Set(banned);
const counts = new Map();
for (const n of nums) counts.set(n, (counts.get(n) || 0) + 1);
const uniqueNums = [...new Set(nums)].sort((a, b) => a - b);
function backtrack(currXor, remaining) {
if (remaining === 0) return 1;
let count = 0;
for (const num of uniqueNums) {
if (counts.get(num) > 0) {
const nextXor = currXor ^ num;
if (!bannedSet.has(nextXor)) {
counts.set(num, counts.get(num) - 1);
count += backtrack(nextXor, remaining - 1);
counts.set(num, counts.get(num) + 1);
}
}
}
return count;
}
let safeCount = 0;
for (const num of uniqueNums) {
const countsCopy = new Map(counts);
countsCopy.set(num, countsCopy.get(num) - 1);
safeCount += backtrack(0, uniqueNums.length - 1);
countsCopy.set(num, countsCopy.get(num) + 1);
}
return safeCount;
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