C++ Program to find a Kth smallest element by the method of partitioning the array.
1. Implement partitioning to find the Kth smallest number from a dataset of n element.
2. The time complexity of this algorithm is O(n*log(n)).
1. Take the input of the data set.
2. Use the partition algorithm.
3. As we place the pivot at the (k-1)th index it will be the kth smallest number.
3. Exit.
C++ program to find kth smallest element by the method of partitioning the array.
This program is successfully run on Dev-C++ using TDM-GCC 4.9.2 MinGW compiler on a Windows system.
#include<iostream> using namespace std; // Swapping two values. void swap(int *a, int *b) { int temp; temp = *a; *a = *b; *b = temp; } // Partitioning the array on the basis of values at high as pivot value. int CreatePartition(int a[], int low, int high) { int pivot, index, i; index = low; pivot = high; // Getting index of pivot. for(i=low; i < high; i++) { if(a[i] < a[pivot]) { swap(&a[i], &a[index]); index++; } } // Swapping value at high and at the index obtained. swap(&a[pivot], &a[index]); return index; } // Implementing Partition. int Partition(int a[], int low, int high, int k) { int pindex; if(low < high) { // Partitioning array using last element as a pivot. // Recursively implementing partitioning in the direction to place the pivot at (k-1)th pivot. pindex = CreatePartition(a, low, high); if(pindex == k-1) return k-1; else if(pindex > k-1) Partition(a, low, pindex-1, k); else Partition(a, pindex+1, high, k); } } int main() { int n, i, k, kk; cout<<"\nEnter the number of data element: "; cin>>n; int arr[n]; for(i = 0; i < n; i++) { cout<<"Enter element "<<i+1<<": "; cin>>arr[i]; } cout<<"\nEnter the k for the kth smallest element: "; cin>>k; kk = Partition(arr, 0, n-1, k); // Printing the result. cout<<"\nThe kth smallest element: "<<arr[kk]; return 0; }
1. Take the input of the data set.
2. Take the input of the value of k.
3. Call Partition().
4. Inside Partition(), rearrange the array according to the pivot using CreatePartition().
5. Get the new index of the pivot as ‘pindex’ and compare it with (k-1).
6. If both the values are equal then return the value as a result to the main().
7. If pindex > k-1, recursively call Partition() for the part before the pivot value.
8. If pindex < k-1, recursively call Partition() for the part after the pivot value.
9. Inside main(), print the kth smallest element.
10. Exit.
Case 1: Enter the number of data element: 10 Enter element 1: 9 Enter element 2: 4 Enter element 3: 0 Enter element 4: 3 Enter element 5: 7 Enter element 6: 8 Enter element 7: 1 Enter element 8: 5 Enter element 9: 6 Enter element 10: 2 Enter the k for the kth smallest element: 4 The kth smallest element: 3
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