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;
        }
    }
}

All Problems

All Solutions