Welcome to Subscribe On Youtube

2598. Smallest Missing Non-negative Integer After Operations

Description

You are given a 0-indexed integer array nums and an integer value.

In one operation, you can add or subtract value from any element of nums.

  • For example, if nums = [1,2,3] and value = 2, you can choose to subtract value from nums[0] to make nums = [-1,2,3].

The MEX (minimum excluded) of an array is the smallest missing non-negative integer in it.

  • For example, the MEX of [-1,2,3] is 0 while the MEX of [1,0,3] is 2.

Return the maximum MEX of nums after applying the mentioned operation any number of times.

 

Example 1:

 Input: nums = [1,-10,7,13,6,8], value = 5 Output: 4 Explanation: One can achieve this result by applying the following operations: - Add value to nums[1] twice to make nums = [1,0,7,13,6,8] - Subtract value from nums[2] once to make nums = [1,0,2,13,6,8] - Subtract value from nums[3] twice to make nums = [1,0,2,3,6,8] The MEX of nums is 4. It can be shown that 4 is the maximum MEX we can achieve.

Example 2:

 Input: nums = [1,-10,7,13,6,8], value = 7 Output: 2 Explanation: One can achieve this result by applying the following operation: - subtract value from nums[2] once to make nums = [1,-10,0,13,6,8] The MEX of nums is 2. It can be shown that 2 is the maximum MEX we can achieve.

 

Constraints:

  • 1 <= nums.length, value <= 105
  • -109 <= nums[i] <= 109

Solutions

Solution 1: Count

We use a hash table or array $cnt$ to count the number of times each remainder of $value$ is taken modulo in the array.

Then start from $0$ and traverse, for the current number $i$ traversed, if $cnt[i \bmod value]$ is $0$, it means that there is no number in the array that takes $i$ modulo $value$ as the remainder, then $i$ is the MEX of the array, and return directly. Otherwise, reduce $cnt[i \bmod value]$ by $1$ and continue to traverse.

The time complexity is $O(n)$ and the space complexity is $O(value)$. Where $n$ is the length of the array $nums$.

  • class Solution { public int findSmallestInteger(int[] nums, int value) { int[] cnt = new int[value]; for (int x : nums) { ++cnt[(x % value + value) % value]; } for (int i = 0;; ++i) { if (cnt[i % value]-- == 0) { return i; } } } } 
  • class Solution { public: int findSmallestInteger(vector<int>& nums, int value) { int cnt[value]; memset(cnt, 0, sizeof(cnt)); for (int x : nums) { ++cnt[(x % value + value) % value]; } for (int i = 0;; ++i) { if (cnt[i % value]-- == 0) { return i; } } } }; 
  • class Solution: def findSmallestInteger(self, nums: List[int], value: int) -> int: cnt = Counter(x % value for x in nums) for i in range(len(nums) + 1): if cnt[i % value] == 0: return i cnt[i % value] -= 1 
  • func findSmallestInteger(nums []int, value int) int { cnt := make([]int, value) for _, x := range nums { cnt[(x%value+value)%value]++ } for i := 0; ; i++ { if cnt[i%value] == 0 { return i } cnt[i%value]-- } } 
  • function findSmallestInteger(nums: number[], value: number): number { const cnt: number[] = new Array(value).fill(0); for (const x of nums) { ++cnt[((x % value) + value) % value]; } for (let i = 0; ; ++i) { if (cnt[i % value]-- === 0) { return i; } } } 
  • /** * @param {number[]} nums * @param {number} value * @return {number} */ var findSmallestInteger = function (nums, value) { const cnt = Array(value).fill(0); for (const x of nums) { ++cnt[((x % value) + value) % value]; } for (let i = 0; ; ++i) { if (cnt[i % value]-- === 0) { return i; } } }; 
  • impl Solution { pub fn find_smallest_integer(nums: Vec<i32>, value: i32) -> i32 { let mut cnt = vec![0; value as usize]; for &x in &nums { let idx = ((x % value + value) % value) as usize; cnt[idx] += 1; } let mut i = 0; loop { let idx = (i % value) as usize; if cnt[idx] == 0 { return i; } cnt[idx] -= 1; i += 1; } } }

All Problems

All Solutions