Given an array arr[] and an integer K. The task is to sort the elements that are in between any two multiples of K.
Examples:
Input: arr[] = {2, 1, 13, 3, 7, 8, 21, 13, 12}, K = 2
Output: 2 1 3 7 13 8 13 21 12
The multiples of 2 in the array are 2, 8 and 12.
The elements that are in between the first two multiples of 2 are 1, 13, 3 and 7.
Hence these elements in sorted order are 1, 3, 7 and 13.
Similarly, the elements between 8 and 12 in sorted order will be 13 and 21.Input: arr[] = {11, 10, 9, 7, 4, 5, 12, 22, 13, 15, 17, 16}, K = 3
Output: 11 10 9 4 5 7 12 13 22 15 17 16
Approach: Traverse the array and keep track of the multiples of K, starting from the 2nd multiple of K sort every element between the current and the previous multiple of K. Print the updated array in the end.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include<bits/stdc++.h> using namespace std; // Utility function to print // the contents of an array void printArr(int arr[], int n) { for (int i = 0; i < n; i++) cout << (arr[i]) << " "; } // Function to sort elements // in between multiples of k void sortArr(int arr[], int n, int k) { // To store the index of // previous multiple of k int prev = -1; for (int i = 0; i < n; i++) { if (arr[i] % k == 0) { // If it is not the // first multiple of k if (prev != -1) // Sort the elements in between // the previous and the current // multiple of k sort(arr + prev + 1, arr + i); // Update previous to be current prev = i; } } // Print the updated array printArr(arr, n); } // Driver code int main() { int arr[] = {2, 1, 13, 3, 7, 8, 21, 13, 12}; int n = sizeof(arr) / sizeof(arr[0]); int k = 2; sortArr(arr, n, k); } // This code is contributed by // Surendra_Gangwar |
Java
// Java implementation of the approach import java.util.Arrays; class GFG { // Utility function to print // the contents of an array static void printArr(int arr[], int n) { for (int i = 0; i < n; i++) System.out.print(arr[i] + " "); } // Function to sort elements // in between multiples of k static void sortArr(int arr[], int n, int k) { // To store the index of // previous multiple of k int prev = -1; for (int i = 0; i < n; i++) { if (arr[i] % k == 0) { // If it is not the // first multiple of k if (prev != -1) // Sort the elements in between // the previous and the current // multiple of k Arrays.sort(arr, prev + 1, i); // Update previous to be current prev = i; } } // Print the updated array printArr(arr, n); } // Driver code public static void main(String[] args) { int arr[] = { 2, 1, 13, 3, 7, 8, 21, 13, 12 }; int n = arr.length; int k = 2; sortArr(arr, n, k); } } |
Python3
# Python3 implementation of the approach # Utility function to print # the contents of an array def printArr(arr, n) : for i in range(n) : print(arr[i], end = " "); # Function to sort elements # in between multiples of k def sortArr(arr, n, k) : # To store the index of # previous multiple of k prev = -1; for i in range(n) : if (arr[i] % k == 0) : # If it is not the first # multiple of k if (prev != -1) : # Sort the elements in between #the previous and the current # multiple of k temp = arr[prev + 1:i]; temp.sort(); arr = arr[ : prev + 1] + temp + arr[i : ]; # Update previous to be current prev = i; # Print the updated array printArr(arr, n); # Driver code if __name__ == "__main__" : arr = [ 2, 1, 13, 3, 7, 8, 21, 13, 12 ]; n = len(arr); k = 2; sortArr(arr, n, k); # This code is contributed by Ryuga |
C#
// C# implementation of the approach using System.Collections; using System; class GFG { // Utility function to print // the contents of an array static void printArr(int []arr, int n) { for (int i = 0; i < n; i++) Console.Write(arr[i] + " "); } // Function to sort elements // in between multiples of k static void sortArr(int []arr, int n, int k) { // To store the index of // previous multiple of k int prev = -1; for (int i = 0; i < n; i++) { if (arr[i] % k == 0) { // If it is not the // first multiple of k if (prev != -1) // Sort the elements in between // the previous and the current // multiple of k Array.Sort(arr, prev + 1, i-(prev + 1)); // Update previous to be current prev = i; } } // Print the updated array printArr(arr, n); } // Driver code public static void Main(String []args) { int []arr = { 2, 1, 13, 3, 7, 8, 21, 13, 12 }; int n = arr.Length; int k = 2; sortArr(arr, n, k); } } //contributed by Arnab Kundu |
2 1 3 7 13 8 13 21 12
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