2661. Smallest Missing Non Negative Integer After Operations¶
Difficulty: Medium
LeetCode Problem View on GitHub
2661. Smallest Missing Non-negative Integer After Operations
Medium
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]andvalue = 2, you can choose to subtractvaluefromnums[0]to makenums = [-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]is0while the MEX of[1,0,3]is2.
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
Solution¶
class Solution {
public int findSmallestInteger(int[] nums, int value) {
HashMap<Integer, Integer> map = new HashMap<>();
for (int ele : nums) {
int curr = ((ele % value) + value) % value;
map.put(curr, map.getOrDefault(curr, 0) + 1);
}
int res = 0;
while (map.getOrDefault(res % value, 0) > 0) {
map.put(res % value, map.getOrDefault(res % value, 0) - 1);
res++;
}
return res;
}
}
Complexity Analysis¶
- Time Complexity:
O(?) - Space Complexity:
O(?)
Approach¶
Detailed explanation of the approach will be added here