76 lines
2.2 KiB
Java
76 lines
2.2 KiB
Java
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//Given two sorted arrays arr1 and arr2 of size M and N respectively and an element K.
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//The task is to find the element that would be at the k’th position of the final sorted array.
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// Time Complexity: O(k)
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// Auxiliary Space: O(1)
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import java.util.*;
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import java.lang.*;
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import java.io.*;
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class Solution {
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public long kthElement( int arr1[], int arr2[], int n, int m, int k) {
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int i=0, j=0, curr_k=0;
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while(i<n && j<m){
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if(arr1[i]<arr2[j]){
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curr_k++;
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if(k==curr_k) return arr1[i];
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i++;
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} else {
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curr_k++;
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if(k==curr_k) return arr2[j];
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j++;
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}
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}
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while (i < n) {
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curr_k++;
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if (k == curr_k)
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return arr1[i];
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i++;
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}
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while (j < m) {
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curr_k++;
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if (k == curr_k)
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return arr2[j];
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j++;
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}
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return -1;
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}
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}
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class Main {
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public static void main(String[] args) throws IOException {
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BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
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int t = Integer.parseInt(br.readLine().trim());
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while(t-->0) {
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StringTokenizer stt = new StringTokenizer(br.readLine());
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int n = Integer.parseInt(stt.nextToken());
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int m = Integer.parseInt(stt.nextToken());
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int k = Integer.parseInt(stt.nextToken());
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int a[] = new int[(int)(n)];
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int b[] = new int[(int)(m)];
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String inputLine[] = br.readLine().trim().split(" ");
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for (int i = 0; i < n; i++) {
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a[i] = Integer.parseInt(inputLine[i]);
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}
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String inputLine1[] = br.readLine().trim().split(" ");
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for (int i = 0; i < m; i++) {
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b[i] = Integer.parseInt(inputLine1[i]);
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}
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Solution obj = new Solution();
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System.out.println(obj.kthElement( a, b, n, m, k));
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}
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}
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}
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/* Test Case:
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Input:
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arr1[] = {2, 3, 6, 7, 9}
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arr2[] = {1, 4, 8, 10}
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k = 5
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Output:
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6
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Explanation:
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The final sorted array would be -
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1, 2, 3, 4, 6, 7, 8, 9, 10
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The 5th element of this array is 6. */
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