77 lines
2.5 KiB
Java
77 lines
2.5 KiB
Java
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/*This is a java program to find kth smallest element form the given sequence of numbers.
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This could be solved by using Quick sort algorithm, where we partition around the pivot element,
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the entire sequence of numbers is broken down to two, we arrange the number such that numbers smaller than pivot is kept in the first sequence and
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numbers larger than the pivot is kept in the second sequence. During this comparison we find the kth smallest element.*/
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//This is a java program to find kth smallest element form the randomly generated sequence using partitioning
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import java.util.Random;
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import java.util.Scanner;
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public class Kth_Smallest_Partitioning
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{
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public static int N = 20;
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public static int[] A = new int[N];
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public static void swap(int dex1, int dex2)
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{
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int temp = A[dex1];
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A[dex1] = A[dex2];
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A[dex2] = temp;
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}
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public static int partition(int start, int end)
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{
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int i = start + 1;
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int j = i;
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int pivot = start;
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for (; i < end; i++)
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{
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if (A[i] < A[pivot])
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{
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swap(i, j);
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j++;
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}
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}
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if (j <= end)
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swap(pivot, (j - 1));
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return j - 1;
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}
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public static void quick_sort(int start, int end, int K)
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{
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int part;
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if (start < end)
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{
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part = partition(start, end);
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if (part == K - 1)
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System.out.println("kth smallest element : " + A[part]);
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if (part > K - 1)
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quick_sort(start, part, K);
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else
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quick_sort(part + 1, end, K);
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}
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return;
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}
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public static void main(String args[])
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{
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Random random = new Random();
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for (int i = 0; i < N; i++)
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A[i] = random.nextInt(1000);
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System.out.println("The original sequence is: ");
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for (int i = 0; i < N; i++)
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System.out.print(A[i] + " ");
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Scanner sc = new Scanner(System.in);
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System.out.println("\nEnter the Kth smallest you want to find: ");
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int k = sc.nextInt();
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quick_sort(0, N, k);
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sc.close();
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}
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}
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/*
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The original sequence is:
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811 30 934 118 942 89 855 917 474 194 630 887 916 997 851 550 917 841 343 202
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Enter the Kth smallest you want to find:
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3
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kth smallest element : 118
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