Given two sorted arrays of size m and n respectively, you are tasked with finding the element that would be at the k’th position of the final sorted array.
Examples:
Input : Array 1 - 2 3 6 7 9
Array 2 - 1 4 8 10
k = 5
Output : 6
Explanation: The final sorted array would be -
1, 2, 3, 4, 6, 7, 8, 9, 10
The 5th element of this array is 6.
Input : Array 1 - 100 112 256 349 770
Array 2 - 72 86 113 119 265 445 892
k = 7
Output : 256
Explanation: Final sorted array is -
72, 86, 100, 112, 113, 119, 256, 265, 349, 445, 770, 892
7th element of this array is 256.
Basic Approach
Since we are given two sorted arrays, we can use merging technique to get the final merged array. From this, we simply go to the k’th index.
C++
// Program to find kth element from two sorted arrays #include <iostream> using namespace std; int kth(int arr1[], int arr2[], int m, int n, int k) { int sorted1[m + n]; int i = 0, j = 0, d = 0; while (i < m && j < n) { if (arr1[i] < arr2[j]) sorted1[d++] = arr1[i++]; else sorted1[d++] = arr2[j++]; } while (i < m) sorted1[d++] = arr1[i++]; while (j < n) sorted1[d++] = arr2[j++]; return sorted1[k - 1]; } int main() { int arr1[5] = {2, 3, 6, 7, 9}; int arr2[4] = {1, 4, 8, 10}; int k = 5; cout << kth(arr1, arr2, 5, 4, k); return 0; } |
Java
// Java Program to find kth element // from two sorted arrays class Main { static int kth(int arr1[], int arr2[], int m, int n, int k) { int[] sorted1 = new int[m + n]; int i = 0, j = 0, d = 0; while (i < m && j < n) { if (arr1[i] < arr2[j]) sorted1[d++] = arr1[i++]; else sorted1[d++] = arr2[j++]; } while (i < m) sorted1[d++] = arr1[i++]; while (j < n) sorted1[d++] = arr2[j++]; return sorted1[k - 1]; } // main function public static void main (String[] args) { int arr1[] = {2, 3, 6, 7, 9}; int arr2[] = {1, 4, 8, 10}; int k = 5; System.out.print(kth(arr1, arr2, 5, 4, k)); } } /* This code is contributed by Harsh Agarwal */ |
Python3
# Program to find kth element # from two sorted arrays def kth(arr1, arr2, m, n, k): sorted1 = [0] * (m + n) i = 0 j = 0 d = 0 while (i < m and j < n): if (arr1[i] < arr2[j]): sorted1[d] = arr1[i] i += 1 else: sorted1[d] = arr2[j] j += 1 d += 1 while (i < m): sorted1[d] = arr1[i] d += 1 i += 1 while (j < n): sorted1[d] = arr2[j] d += 1 j += 1 return sorted1[k - 1] # driver code arr1 = [2, 3, 6, 7, 9] arr2 = [1, 4, 8, 10] k = 5; print(kth(arr1, arr2, 5, 4, k)) # This code is contributed by Smitha Dinesh Semwal |
C#
// C# Program to find kth element // from two sorted arrays class GFG { static int kth(int[] arr1, int[] arr2, int m, int n, int k) { int[] sorted1 = new int[m + n]; int i = 0, j = 0, d = 0; while (i < m && j < n) { if (arr1[i] < arr2[j]) sorted1[d++] = arr1[i++]; else sorted1[d++] = arr2[j++]; } while (i < m) sorted1[d++] = arr1[i++]; while (j < n) sorted1[d++] = arr2[j++]; return sorted1[k - 1]; } // Driver Code static void Main() { int[] arr1 = {2, 3, 6, 7, 9}; int[] arr2 = {1, 4, 8, 10}; int k = 5; System.Console.WriteLine(kth(arr1, arr2, 5, 4, k)); } } // This code is contributed by mits |
PHP
<?php // Program to find kth // element from two // sorted arrays function kth($arr1, $arr2, $m, $n, $k) { $sorted1[$m + $n] = 0; $i = 0; $j = 0; $d = 0; while ($i < $m && $j < $n) { if ($arr1[$i] < $arr2[$j]) $sorted1[$d++] = $arr1[$i++]; else $sorted1[$d++] = $arr2[$j++]; } while ($i < $m) $sorted1[$d++] = $arr1[$i++]; while ($j < $n) $sorted1[$d++] = $arr2[$j++]; return $sorted1[$k - 1]; } // Driver Code $arr1 = array(2, 3, 6, 7, 9); $arr2 = array(1, 4, 8, 10); $k = 5; echo kth($arr1, $arr2, 5, 4, $k); // This code is contributed // by ChitraNayal ?> |
Output:
6
Time Complexity: O(n)
Auxiliary Space : O(m + n)
Divide And Conquer Approach 1
While the previous method works, can we make our algorithm more efficient? The answer is yes. By using a divide and conquer approach, similar to the one used in binary search, we can attempt to find the k’th element in a more efficient way.
Explanation: We compare the middle elements of arrays arr1 and arr2, let us call these indices mid1 and mid2 respectively. Let us assume arr1[mid1] k, then clearly the elements after mid2 cannot be the required element. We then set the last element of arr2 to be arr2[mid2]. In this way, we define a new subproblem with half the size of one of the arrays.
C++
// Program to find k-th element from two sorted arrays #include <iostream> using namespace std; int kth(int *arr1, int *arr2, int *end1, int *end2, int k) { if (arr1 == end1) return arr2[k]; if (arr2 == end2) return arr1[k]; int mid1 = (end1 - arr1) / 2; int mid2 = (end2 - arr2) / 2; if (mid1 + mid2 < k) { if (arr1[mid1] > arr2[mid2]) return kth(arr1, arr2 + mid2 + 1, end1, end2, k - mid2 - 1); else return kth(arr1 + mid1 + 1, arr2, end1, end2, k - mid1 - 1); } else { if (arr1[mid1] > arr2[mid2]) return kth(arr1, arr2, arr1 + mid1, end2, k); else return kth(arr1, arr2, end1, arr2 + mid2, k); } } int main() { int arr1[5] = {2, 3, 6, 7, 9}; int arr2[4] = {1, 4, 8, 10}; int k = 5; cout << kth(arr1, arr2, arr1 + 5, arr2 + 4, k - 1); return 0; } |
Output:
6
Note that in the above code, k is 0 indexed, which means if we want a k that’s 1 indexed, we have to subtract 1 when passing it to the function.
Time Complexity: O(log n + log m)
Divide And Conquer Approach 2
While the above implementation is very efficient, we can still get away with making it more efficient. Instead of dividing the array into segments of n / 2 and m / 2 then recursing, we can divide them both by k / 2 and recurse. Below implementation displays this.
Explanation: Instead of comparing the middle element of the arrays, we compare the k / 2th element. Let arr1 and arr2 be the arrays. Now, if arr1[k / 2] arr1[1] New subproblem: Array 1 - 6 7 9 Array 2 - 1 4 8 10 k = 5 - 2 = 3 floor(k / 2) = 1 arr1[1] = 6 arr2[1] = 1 arr1[1] > arr2[1] New subproblem: Array 1 - 6 7 9 Array 2 - 4 8 10 k = 3 - 1 = 2 floor(k / 2) = 1 arr1[1] = 6 arr2[1] = 4 arr1[1] > arr2[1] New subproblem: Array 1 - 6 7 9 Array 2 - 8 10 k = 2 - 1 = 1 Now, we directly compare first elements, since k = 1. arr1[1] < arr2[1] Hence, arr1[1] = 6 is the answer.
C++
// Program to find kth element from two sorted arrays #include <iostream> using namespace std; int kth(int arr1[], int arr2[], int m, int n, int k, int st1 = 0, int st2 = 0) { // In case we have reached end of array 1 if (st1 == m) return arr2[st2 + k - 1]; // In case we have reached end of array 2 if (st2 == n) return arr1[st1 + k - 1]; // k should never reach 0 or exceed sizes // of arrays if (k == 0 || k > (m - st1) + (n - st2)) return -1; // Compare first elements of arrays and return if (k == 1) return (arr1[st1] < arr2[st2]) ? arr1[st1] : arr2[st2]; int curr = k / 2; // Size of array 1 is less than k / 2 if (curr - 1 >= m - st1) { // Last element of array 1 is not kth // We can directly return the (k - m)th // element in array 2 if (arr1[m - 1] < arr2[st2 + curr - 1]) return arr2[st2 + (k - (m - st1) - 1)]; else return kth(arr1, arr2, m, n, k - curr, st1, st2 + curr); } // Size of array 2 is less than k / 2 if (curr-1 >= n-st2) { if (arr2[n - 1] < arr1[st1 + curr - 1]) return arr1[st1 + (k - (n - st2) - 1)]; else return kth(arr1, arr2, m, n, k - curr, st1 + curr, st2); } else { // Normal comparison, move starting index // of one array k / 2 to the right if (arr1[curr + st1 - 1] < arr2[curr + st2 - 1]) return kth(arr1, arr2, m, n, k - curr, st1 + curr, st2); else return kth(arr1, arr2, m, n, k - curr, st1, st2 + curr); } } // Driver code int main() { int arr1[5] = {2, 3, 6, 7, 9}; int arr2[4] = {1, 4, 8, 10}; int k = 5; cout << kth(arr1, arr2, 5, 4, k); return 0; } |
Output:
6
Time Complexity: O(log k)
Now, k can take a maximum value of m + n. This means that log k can be in the worst case, log(m + n). Logm + logn = log(mn) by properties of logarithms, and when m, n > 2, log(m + n) < log(mn). Thus this algorithm slightly outperforms the previous algorithm.Also see another simple implemented log k approach suggested by Raj Kumar.
C++
// C++ Program to find kth element from two sorted arrays // Time Complexity: O(log k) #include <iostream> using namespace std; int kth(int arr1[], int m, int arr2[], int n, int k) { if (k > (m+n) || k < 1) return -1; // let m <= n if (m > n) return kth(arr2, n, arr1, m, k); // if arr1 is empty returning k-th element of arr2 if (m == 0) return arr2[k - 1]; // if k = 1 return minimum of first two elements of both arrays if (k == 1) return min(arr1[0], arr2[0]); // now the divide and conquer part int i = min(m, k / 2), j = min(n, k / 2); if (arr1[i - 1] > arr2[j - 1] ) // Now we need to find only k-j th element since we have found out the lowest j return kth(arr1, m, arr2 + j, n - j, k - j); else // Now we need to find only k-i th element since we have found out the lowest i return kth(arr1 + i, m - i, arr2, n, k - i); } // Driver code int main() { int arr1[5] = {2, 3, 6, 7, 9}; int arr2[4] = {1, 4, 8, 10}; int m = sizeof(arr1)/sizeof(arr1[0]); int n = sizeof(arr2)/sizeof(arr2[0]); int k = 5; int ans = kth(arr1,m,arr2, n, k); if(ans == -1) cout<<"Invalid query"; else cout<<ans; return 0; } // This code is contributed by Raj Kumar |
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