Check if all array elements can be converted to K using given operations

Last Updated : 19 Jul, 2021

Given an integer array arr of size N and an integer K, the task is to make all elements of the array equal to K using the following operations:

Choose an arbitrary subarray [l….r] of the input array
Replace all values of this subarray equal to the [((r – l) + 2) / 2]^th value in sorted subarray [l…r]

Examples:

Input: arr [ ] = {5, 4, 3, 1, 2, 6, 7, 8, 9, 10}, K = 3
Output: YES
Explanation:
Choose first five elements and replace all elements with 3: arr = {3, 3, 3, 3, 3, 6, 7, 8, 9, 10}
Now take forth to sixth elements and replace all elements with 3: arr = {3, 3, 3, 3, 3, 3, 7, 8, 9, 10}
By applying same operations we can make all elements of arr equal to K: arr = {3, 3, 3, 3, 3, 3, 3, 3, 3, 3}
Input: arr [ ] = {3, 1, 2, 3}, K = 4
Output: NO

Approach:

We can observe that it is possible to make all elements of the array equal to K only if following two conditions are satisfied:
1. There must be at least one element equal to K.
2. There must exist a continuous triplet such that any two values of that triplet is greater than or equal to K.
To solve this problem, we need to create an auxiliary array, say aux[] , that contains three values 0, 1, 2.
The final task is to check if it is possible to make all elements of aux array equal to 1. If two out of three continuous elements in aux[] are greater than 0, then we can take a subarray of size 3 and make all elements of that subarray equal to 1. And then we expand this logic through the entire array.

Below is the implementation of above approach:

C++

   // C++ implementation of above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function that prints
// whether is to possible
// to make all elements
// of the array equal to K
void makeAllK(int a[], int k, int n)
{
    vector<int> aux;
 
    bool one_found = false;
 
    // Fill vector aux
    // according to the
    // above approach
    for (int i = 0; i < n; i++) {
 
        if (a[i] < k)
            aux.push_back(0);
 
        else if (a[i] == k) {
            aux.push_back(1);
            one_found = true;
        }
 
        else
            aux.push_back(2);
    }
 
    // Condition if K
    // does not exist in
    // the given array
    if (one_found == false) {
        cout << "NO"
             << "\n";
        return;
    }
 
    bool ans = false;
 
    if (n == 1
        && aux[0] == 1)
        ans = true;
 
    if (n == 2
        && aux[0] > 0
        && aux[1] > 1)
        ans = true;
 
    for (int i = 0; i < n - 2; i++) {
 
        // Condition for minimum
        // two elements is
        // greater than 0 in
        // pair of three elements
        if (aux[i] > 0
            && aux[i + 1] > 0) {
 
            ans = true;
            break;
        }
 
        else if (aux[i] > 0
                 && aux[i + 2] > 0) {
            ans = true;
            break;
        }
 
        else if (aux[i + 2] > 0
                 && aux[i + 1] > 0) {
            ans = true;
            break;
        }
    }
 
    if (ans == true)
        cout << "YES"
             << "\n";
    else
        cout << "NO"
             << "\n";
}
 
// Driver Code
int main()
{
    int arr[]
        = { 1, 2, 3,
            4, 5, 6,
            7, 8, 9, 10 };
 
    int K = 3;
 
    int size = sizeof(arr)
               / sizeof(arr[0]);
 
    makeAllK(arr, K, size);
 
    return 0;
}
 
  

Java

   // Java implementation of the above approach
import java.util.*;
 
class GFG{
 
// Function that prints
// whether is to possible
// to make all elements
// of the array equal to K
static void makeAllK(int a[], int k, int n)
{
    Vector<Integer> aux = new Vector<Integer>();
 
    boolean one_found = false;
 
    // Fill vector aux according 
    // to the above approach
    for(int i = 0; i < n; i++)
    {
       if (a[i] < k)
           aux.add(0);
        
       else if (a[i] == k) 
       {
           aux.add(1);
           one_found = true;
       }
       else
           aux.add(2);
    }
 
    // Condition if K does not  
    // exist in the given array
    if (one_found == false)
    {
        System.out.print("NO" + "\n");
        return;
    }
 
    boolean ans = false;
 
    if (n == 1 && aux.get(0) == 1)
        ans = true;
 
    if (n == 2 && aux.get(0) > 0 && 
                  aux.get(1) > 1)
        ans = true;
 
    for(int i = 0; i < n - 2; i++)
    {
        
       // Condition for minimum
       // two elements is
       // greater than 0 in
       // pair of three elements
       if (aux.get(i) > 0 && 
           aux.get(i + 1) > 0)
       {
           ans = true;
           break;
       }
       else if (aux.get(i) > 0 &&
                aux.get(i + 2) > 0) 
       {
           ans = true;
           break;
       }
       else if (aux.get(i + 2) > 0 && 
                aux.get(i + 1) > 0)
       {
           ans = true;
           break;
       }
    }
 
    if (ans == true)
        System.out.print("YES" + "\n");
    else
        System.out.print("NO" + "\n");
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5,
                  6, 7, 8, 9, 10 };
    int K = 3;
    int size = arr.length;
 
    makeAllK(arr, K, size);
 
}
}
 
// This code is contributed by amal kumar choubey
 
  

Python3

   # Python3 implementation of above approach
 
# Function that prints whether is
# to possible to make all elements
# of the array equal to K
def makeAllK(a, k, n):
     
    aux = []
    one_found = False
 
    # Fill vector aux according
    # to the above approach
    for i in range(n):
        if (a[i] < k):
            aux.append(0)
 
        elif (a[i] == k):
            aux.append(1)
            one_found = True
 
        else:
            aux.append(2)
 
    # Condition if K does 
    # not exist in the given
    # array
    if (one_found == False):
        print("NO")
        return
 
    ans = False
 
    if (n == 1 and aux[0] == 1):
        ans = True
 
    if (n == 2 and aux[0] > 0 and aux[1] > 1):
        ans = True
 
    for i in range(n - 2):
         
        # Condition for minimum two 
        # elements is greater than 
        # 0 in pair of three elements
        if (aux[i] > 0 and aux[i + 1] > 0):
            ans = True
            break
 
        elif (aux[i] > 0 and aux[i + 2] > 0):
            ans = True
            break
 
        elif (aux[i + 2] > 0 and aux[i + 1] > 0):
            ans = True
            break
 
    if (ans == True):
        print("YES")
    else:
        print("NO")
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
    K = 3
    size = len(arr)
 
    makeAllK(arr, K, size)
 
# This code is contributed by Surendra_Gangwar
 
  

C#

   // C# implementation of the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function that prints
// whether is to possible
// to make all elements
// of the array equal to K
static void makeAllK(int []a, int k, int n)
{
    List<int> aux = new List<int>();
 
    bool one_found = false;
 
    // Fill vector aux according 
    // to the above approach
    for(int i = 0; i < n; i++)
    {
       if (a[i] < k)
       {
           aux.Add(0);
       }
       else if (a[i] == k) 
       {
           aux.Add(1);
           one_found = true;
       }
       else
           aux.Add(2);
    }
 
    // Condition if K does not 
    // exist in the given array
    if (one_found == false)
    {
        Console.Write("NO" + "\n");
        return;
    }
 
    bool ans = false;
    if (n == 1 && aux[0] == 1)
        ans = true;
 
    if (n == 2 && aux[0] > 0 && 
                  aux[1] > 1)
        ans = true;
 
    for(int i = 0; i < n - 2; i++)
    {
        
       // Condition for minimum
       // two elements is
       // greater than 0 in
       // pair of three elements
       if (aux[i] > 0 && 
           aux[i + 1] > 0)
       {
           ans = true;
           break;
       }
       else if (aux[i] > 0 && 
                aux[i + 2] > 0) 
       {
           ans = true;
           break;
       }
       else if (aux[i + 2] > 0 && 
                aux[i + 1] > 0)
       {
           ans = true;
           break;
       }
    }
    if (ans == true)
        Console.Write("YES" + "\n");
    else
        Console.Write("NO" + "\n");
}
 
// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5,
                  6, 7, 8, 9, 10 };
    int K = 3;
    int size = arr.Length;
 
    makeAllK(arr, K, size);
}
}
 
// This code is contributed by amal kumar choubey
 
  

Javascript

   <script>
 
// Javascript implementation of above approach
 
// Function that prints
// whether is to possible
// to make all elements
// of the array equal to K
function makeAllK(a, k, n)
{
    var aux = [];
 
    var one_found = false;
 
    // Fill vector aux
    // according to the
    // above approach
    for (var i = 0; i < n; i++) {
 
        if (a[i] < k)
            aux.push(0);
 
        else if (a[i] == k) {
            aux.push(1);
            one_found = true;
        }
 
        else
            aux.push(2);
    }
 
    // Condition if K
    // does not exist in
    // the given array
    if (one_found == false) {
        document.write( "NO" + "<br>");
        return;
    }
 
    var ans = false;
 
    if (n == 1
        && aux[0] == 1)
        ans = true;
 
    if (n == 2
        && aux[0] > 0
        && aux[1] > 1)
        ans = true;
 
    for (var i = 0; i < n - 2; i++) {
 
        // Condition for minimum
        // two elements is
        // greater than 0 in
        // pair of three elements
        if (aux[i] > 0
            && aux[i + 1] > 0) {
 
            ans = true;
            break;
        }
 
        else if (aux[i] > 0
                 && aux[i + 2] > 0) {
            ans = true;
            break;
        }
 
        else if (aux[i + 2] > 0
                 && aux[i + 1] > 0) {
            ans = true;
            break;
        }
    }
 
    if (ans == true)
        document.write( "YES"+ "<br>")
    else
        document.write( "NO"+ "<br>")
}
 
// Driver Code
var arr
    = [1, 2, 3,
        4, 5, 6,
        7, 8, 9, 10];
var K = 3;
var size = arr.length;
makeAllK(arr, K, size);
 
// This code is contributed by rutvik_56.
</script>
 
  

Output:

YES

Time Complexity: O(N)
Auxiliary Space: O(N)

Check if the Array elements can be made 0 with given conditions

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Check if all array elements can be converted to K using given operations

C++

Java

Python3

C#

Javascript

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