1539. Kth Missing Positive Number
Problem Statement
Given an array arr
of positive integers sorted in a strictly increasing order, and an integer k
.
Return the kth
positive integer that is missing from this array.
Example 1:
Input: arr = [2,3,4,7,11], k = 5
Output: 9
Explanation: The missing positive integers are [1,5,6,8,9,10,12,13,...]. The 5th missing positive integer is 9.
Example 2:
Input: arr = [1,2,3,4], k = 2
Output: 6
Explanation: The missing positive integers are [5,6,7,...]. The 2nd missing positive integer is 6.
Constraints:
1 <= arr.length <= 1000
1 <= arr[i] <= 1000
1 <= k <= 1000
arr[i] < arr[j]
for1 <= i < j <= arr.length
Intuition
Approach:
We try to find smallest range in which an element may lie
Then we kind of find the number
eg->
2 3 4 7 11
msg 1 1 1 3 6
how we found
idealy at position of 11, there should be 5 , So missing is 11-5
we want k=5th mssig
We get low at 11th, high=7th
So extra numbers we require are 2-> at high 3 are misssing for 5 we need 2
Hence 7+(2) = 9
Links
https://leetcode.com/problems/kth-missing-positive-number/description/
Video Links
https://www.youtube.com/watch?v=uZ0N_hZpyps&ab_channel=takeUforward
Approach 1:
n time
class Solution {
public:
int findKthPositive(vector<int>& arr, int k) {
for(auto &it: arr){
if(it > k)
break;
else
k++;
}
return k;
}
};
Approach 2:
logn time
class Solution {
public:
int findKthPositive(vector<int>& arr, int k) {
int low=0, high=arr.size()-1;
while(low<=high){
int mid = low+(high-low)/2;
int missing = arr[mid] - (mid+1);
if(missing >= k) high=mid-1;
else low=mid+1;
}
return high+1+k;
}
};
Approach 3:
Approach 4:
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