Check if Array element can be incremented to X by using given operations

Given an array A[] of size N, and an integer X, the following operation can be performed on that array:

• Choose any element, say Y from the array.
• Then every element in the array except Y are incremented.
• In the next step, only the value Y+1 can be chosen and the steps are repeated until Y+1 is not present in the array.

The task is to return "YES" if Y can be made strictly greater than X, otherwise, return "NO".

Examples : 

Input N = 9, A[] = { 2, 2, 1, 3, 4, 2, 1, 2, 2 }, X = 5
Output: YES
Explanation: Choose Y = 2 at index 0, then A becomes {2, 3, 2, 4, 5, 3, 2, 3, 3}
Then Y can only be 2 +1 = 3, so choose Y = 3 at index 1
Then A becomes {3, 3, 3, 5, 6, 4, 3, 4, 4}
Then Y can only be 3 +1 = 4, so choose Y = 4 at index 6
Then A becomes {4, 4, 4, 6, 7, 5, 4, 4, 5}
Then Y can only be 4 +1 = 5, so choose Y = 5 at index 7.
Then A becomes {5, 5, 5, 7, 8, 6, 5, 5, 5}
Then Y can only be 5 +1 = 6, so choose Y = 6 at index 4.
Which is greater than X (6 > 5), so return YES

Input: N = 4, A[] = {2, 3, 4, 1}, X = 4
Output: NO

Approach: 

It can be observed in this problem that the maximum value that any element of the array a[i] can achieve would be a[i] + freq[a[i]] - 1, where freq[a[i]] is the frequency of the array element, and a hashing data structure can be used to calculate the frequency of elements of the array.

 Follow the steps mentioned below to implement the idea:-

• Create a HashMap and count the frequency of each distinct element by iterating on arr[], and save them in a HashMap.
• Traverse on the map and apply the formula: a[i] + freq[a[i]] - 1 for each pair stored in the map.
• Now, If the maximum value achieved is strictly greater than X,  then print "YES" or else print "NO". 

Below is the implementation of this approach: 

// C++ code implementation
#include <bits/stdc++.h>
using namespace std;

// Function to check whether maximum value
// greater than X can be achieved or not
string find_max(int N, int A[], int X)
{

  // Map initialized for counting Frequency
  unordered_map<int,int> map;

  // Loop to insert values in Map
  for (int i = 0; i < N; i++) {

    // Obtaining frequency of each
    // distinct element in A[]
    int freq = map[A[i]] == 0
      ? 1
      : map[A[i]] + 1;

    // Putting in Map
    map[A[i]]=freq;
  }

  // Variable to store maximum value achieved
  long max_value = 0;

  // Loop for iterating over Map through Entry Set
  for (auto it:map) {

    // Finding Current maximum value achieved by
    // using formula: (element+(frequency-1))
    int current
      = it.first + (it.second - 1);

    // Updating max_value if current is greater
    // than max_value
    max_value
      = current > max_value ? current : max_value;
  }

  // Printing YES/NO based on comparison
  // between maximum value and X.
  return max_value > X ? "YES" : "NO";
}

int main()
{
  int N = 9;
  int A[] = { 2, 2, 1, 3, 4, 2, 1, 2, 2 };
  int X = 5;
  cout<<(find_max(N, A, X))<<endl;
  return 0;
}

// This code is contributed by ksam24000

// Java code to implement approach

import java.util.*;

class GFG {

    // Function to check whether maximum value
    // greater than X can be achieved or not
    static String find_max(int N, int[] A, int X)
    {
        // Map initialized for counting Frequency
        HashMap<Integer, Integer> map
            = new HashMap<>();

        // Loop to insert values in Map
        for (int i = 0; i < N; i++) {

            // Obtaining frequency of each
            // distinct element in A[]
            int freq = map.get(A[i]) == null
                           ? 1
                           : map.get(A[i]) + 1;

            // Putting in Map
            map.put(A[i], freq);
        }

        // Variable to store maximum value achieved
        long max_value = 0;

        // Loop for iterating over Map through Entry Set
        for (Map.Entry<Integer, Integer> set :
             map.entrySet()) {

            // Finding Current maximum value achieved by
            // using formula: (element+(frequency-1))
            int current
                = set.getKey() + (set.getValue() - 1);

            // Updating max_value if current is greater
            // than max_value
            max_value
                = current > max_value ? current : max_value;
        }

        // Printing YES/NO based on comparison
        // between maximum value and X.
        return max_value > X ? "YES" : "NO";
    }

    // Driver function
    public static void main(String[] args)
    {
        int N = 9;
        int[] A = { 2, 2, 1, 3, 4, 2, 1, 2, 2 };
        int X = 5;

        // Function call
        System.out.println(find_max(N, A, X));
    }
}

# Python code to implement the above approach

# Function to check whether maximum value 
# greater than X can be achieved or not
def find_max(N, A, X):
    # Map initialized for counting frequency
    map = {}

    # loop to insert values in map
    for i in range(N):
        # Obtaining frequency of each 
        # distinct element in A[]
        if A[i] in map:
            map[A[i]] += 1

        # Putting in map
        else:
            map[A[i]] = 1

    # Variable to store maximum value achieved
    max_value = 0

    # loop for iterating over map
    for first, second in map.items():

        # Finding current maximum value achieved by using 
        # formula: (element + (frequency-1))
        current = first + second - 1

        # Updating max_value if current is greater 
        # than max_value
        if current > max_value:
            max_value = current

    # Printing Yes/ No based on comparison 
    # between maximum value and X.
    if max_value > X:
        return "YES"
    else:
        return "NO"

N = 9
A = [2, 2, 1, 3, 4, 2, 1, 2, 2]
X = 5

# Function call
print(find_max(N, A, X))

# This code is contributed by lokesh.

// C# code to implement approach
using System;
using System.Collections.Generic;

public class GFG {

  // Function to check whether maximum value
  // greater than X can be achieved or not
  static String find_max(int N, int[] A, int X)
  {

    // Map initialized for counting Frequency
    Dictionary<int, int> map
      = new Dictionary<int, int>();

    // Loop to insert values in Map
    for (int i = 0; i < N; i++) {
      if (map.ContainsKey(A[i])) {
        map[A[i]] += 1;
      }
      else {
        // Putting in Map
        map.Add(A[i], 1);
      }
    }

    // Variable to store maximum value achieved
    long max_value = 0;

    // Loop for iterating over Map through Entry Set
    foreach(var Set in map)
    {

      // Finding Current maximum value achieved by
      // using formula: (element+(frequency-1))
      int current = Set.Key + (Set.Value - 1);

      // Updating max_value if current is greater
      // than max_value
      max_value
        = current > max_value ? current : max_value;
    }

    // Printing YES/NO based on comparison
    // between maximum value and X.
    return max_value > X ? "YES" : "NO";
  }

  static public void Main()
  {

    // Code
    int N = 9;
    int[] A = { 2, 2, 1, 3, 4, 2, 1, 2, 2 };
    int X = 5;

    // Function call
    Console.WriteLine(find_max(N, A, X));
  }
}

// This code is contributed by lokeshmvs21.

<script>

    // JavaScript code to implement approach

    // Function to check whether maximum value
    // greater than X can be achieved or not
    function find_max(N, A, X){
        
        // Map initialized for counting Frequency
        let map = new Map();
        
        // Loop to insert values in Map
        for(let i=0;i<N;i++){
            // Obtaining frequency of each
            // distinct element in A[]
            if(map.has(A[i])){
                map.set(A[i], map.get(A[i])+1);
            }
            
            // Putting in Map
            else{
                map.set(A[i], 1);
            }
        }
        
        // Variable to store maximum value achieved
        let max_value = 0;
        
         // Loop for iterating over Map through Entry Set
        map.forEach((values, keys)=>{
        
            // Finding Current maximum value achieved by
            // using formula: (element+(frequency-1))
            let current = keys + values - 1;
            
            // Updating max_value if current is greater
            // than max_value
            max_value = current>max_value ? current : max_value;
        })
        
        // Printing YES/NO based on comparison
        // between maximum value and X.
        return (max_value>X) ? "YES" : "NO";
    }

    let N = 9;
    let A = [ 2, 2, 1, 3, 4, 2, 1, 2, 2];
    let X = 5;
    
    // Function call
    document.write(find_max(N, A, X))
    
    // This code is contributed by lokeshmvs21.
</script>

YES

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

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Article Tags :

• Mathematical
• Competitive Programming
• DSA
• Arrays
• frequency-counting

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Practice Tags :

• Arrays
• Mathematical