PROBLEM DESCRIPTION
Given two sorted arrays arr1 and arr2 of size N and M respectively and an element K. The task is to find the element that would be at the kth position of the final sorted array.
SOLUTION
APPROACH 1 | Time Complexity -> O(n+m)
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
class Solution {
public long kthElement(int arr1[], int arr2[], int n, int m, int k) {
// Initialize indices for arr1 and arr2
int i = 0;
int j = 0;
// Continue until kth element is found
while (k > 1) {
// If all elements of arr1 are already used, move to the next element of arr2
if (i >= n) {
j++;
k--;
continue;
}
// If all elements of arr2 are already used, move to the next element of arr1
if (j >= m) {
i++;
k--;
continue;
}
// Compare elements of arr1 and arr2, move to the smaller one
if (arr1[i] < arr2[j]) {
i++;
} else {
j++;
}
// Decrement k since we've considered one element
k--;
}
// If all elements of arr1 are already used, return the element from arr2
if (i >= n)
return arr2[j];
// If all elements of arr2 are already used, return the element from arr1
if (j >= m)
return arr1[i];
// Return the minimum of current elements from arr1 and arr2
return Math.min(arr1[i], arr2[j]);
}
}