Modify given array by incrementing first occurrence of every element by K

Given an array arr[] consisting of N integers, read every element of the array one by one and perform the following operations:

• If the current element arr[i] had previously occurred in the array, increase its first occurrence by K.
• Otherwise, insert arr[i] into the sequence

The task is to print the final sequence of integers obtained by performing the above operations

Examples:

Input: arr[] = {1, 2, 3, 2, 3}, K = 1
Output: [1, 4, 3, 2, 3]
Explanation:

Arrival : 1
Since 1 is the first element in the stream, simply insert it into the solution.
Therefore, b[] = [1]

Arrival: 2
Since 2 is not existing in the array, simply insert it into the solution.
Therefore, b[] = [1, 2]

Arrival: 3
Since 3 is not existing in the array, simply insert it into the solution.
Therefore, b[] = [1, 2, 3]

Arrival: 2
Since 2 already exists, increasing its first occurrence by K(=1)modifies the array b[] to [1, 3, 3, 2]

Arrival: 3
Since 3 already exists, increasing its first occurrence by K(=1)modifies the array b[] to [1, 4, 3, 2, 3]

Input: arr[] = {1, 4, 1, 1, 4}, K = 6
Output: [7, 10, 7, 1, 4]

Naive Approach: The simplest approach to solve the problem is to traverse the array, and for every array element arr[i], traverse in the range [0, i – 1] to check if arr[i] is already present in the array or not. If found to be true, increase the first occurrence of arr[i] by K.

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized using Hashing. Follow the steps below to solve the problem:

• Traverse the array and store the occurrence of every array element in a Map paired with the index of its occurrence in increasing order.
• If arr[i] is found to be already present in the Map, remove the first occurrence of arr[i] from the Map. Insert that index paired with arr[i] + K as the key back into the Map.
• Repeat the above steps for all array elements. Once, the entire array is traversed, obtain the sequence of integers from the Map and print the final sequence.

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach

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

// Print the required final sequence
void printSequence(vector<int>& A,
                   int n, int k)
{
    // Stores the array element-index pairs
    unordered_map<int, set<int> > mp;

    // Stores the required sequence
    vector<int> sol;

    // Insert all array elements
    for (int x : A)
        sol.push_back(x);

    for (int i = 0; i < n; i++) {
        // If current element has
        // not occurred previously
        if (mp.find(sol[i]) == mp.end()
            || mp[sol[i]].size() == 0) {
            mp[sol[i]].insert(i);
        }

        // Otherwise
        else {

            // Iterator to the first index
            // containing sol[i]
            auto idxx = mp[sol[i]].begin();

            int idx = *idxx;

            // Remove that occurrence
            mp[sol[i]].erase(idxx);

            // Increment by K
            sol[idx] += k;

            // Insert the incremented
            // element at that index
            mp[sol[idx]].insert(idx);
            mp[sol[i]].insert(i);
        }
    }

    // Print the final sequence
    for (int x : sol) {
        cout << x << " ";
    }
}

// Driver Code
int main()
{
    int N = 5;
    int K = 6;
    vector<int> A = { 1, 4, 1, 1, 4 };
    printSequence(A, N, K);
}

Java

import java.util.*;

public class Main {

    // Print the required final sequence
    static void printSequence(int[] A, int n, int k) {
        // Stores the array element-index pairs
        Map<Integer, Set<Integer>> mp = new HashMap<>();

        // Stores the required sequence
        List<Integer> sol = new ArrayList<>();

        // Insert all array elements
        for (int x : A)
            sol.add(x);

        for (int i = 0; i < n; i++) {
            // If current element has not occurred previously
            if (!mp.containsKey(sol.get(i)) || mp.get(sol.get(i)).size() == 0) {
                mp.computeIfAbsent(sol.get(i), key -> new HashSet<>()).add(i);
            } else {
                // Iterator to the first index containing sol[i]
                Iterator<Integer> idxx = mp.get(sol.get(i)).iterator();

                int idx = idxx.next();

                // Remove that occurrence
                idxx.remove();

                // Increment by K
                sol.set(idx, sol.get(idx) + k);

                // Insert the incremented element at that index
                mp.computeIfAbsent(sol.get(idx), key -> new HashSet<>()).add(idx);
                mp.computeIfAbsent(sol.get(i), key -> new HashSet<>()).add(i);
            }
        }

        // Print the final sequence
        for (int x : sol) {
            System.out.print(x + " ");
        }
    }

    // Driver Code
    public static void main(String[] args) {
        int N = 5;
        int K = 6;
        int[] A = {1, 4, 1, 1, 4};
        printSequence(A, N, K);
    }
}

Python3

# Python3 Program to implement the above approach

import collections

def printSequence(A, n, k):
    # Stores the array element-index pairs
    mp = collections.defaultdict(set)

    # Stores the required sequence
    sol = []

    # Insert all array elements
    for x in A:
        sol.append(x)

    for i in range(n):
        # If current element has not occurred previously
        if sol[i] not in mp or len(mp[sol[i]]) == 0:
            mp[sol[i]].add(i)
        # Otherwise
        else:
            # Get the first index containing sol[i]
            idx = mp[sol[i]].pop()

            # Increment by K
            sol[idx] += k

            # Insert the incremented element at that index
            mp[sol[idx]].add(idx)
            mp[sol[i]].add(i)

    # Print the final sequence
    print(" ".join(map(str, sol)))

# Driver Code
N = 5
K = 6
A = [1, 4, 1, 1, 4]
printSequence(A, N, K)

C#

using System;
using System.Collections.Generic;

class Program {
  static void Main(string[] args) {
    int N = 5;
    int K = 6;
    List<int> A = new List<int> { 1, 4, 1, 1, 4 };
    printSequence(A, N, K);
  }

  static void printSequence(List<int> A, int n, int k) {
    // Stores the array element-index pairs
    Dictionary<int, HashSet<int>> mp = new Dictionary<int, HashSet<int>>();

    // Stores the required sequence
    List<int> sol = new List<int>();

    // Insert all array elements
    foreach (int x in A)
      sol.Add(x);

    for (int i = 0; i < n; i++) {
      // If current element has not occurred previously
      if (!mp.ContainsKey(sol[i]) || mp[sol[i]].Count == 0) {
        mp[sol[i]] = new HashSet<int>();
        mp[sol[i]].Add(i);
      }

      // Otherwise
      else {
        // Iterator to the first index containing sol[i]
        var idxx = mp[sol[i]].GetEnumerator();
        idxx.MoveNext();
        int idx = idxx.Current;

        // Remove that occurrence
        mp[sol[i]].Remove(idx);

        // Increment by K
        sol[idx] += k;

        // Insert the incremented element at that index
        if (!mp.ContainsKey(sol[idx])) {
          mp[sol[idx]] = new HashSet<int>();
        }
        mp[sol[idx]].Add(idx);
        mp[sol[i]].Add(i);
      }
    }

    // Print the final sequence
    foreach (int x in sol) {
      Console.Write(x + " ");
    }
  }
}

Javascript

<script>

// Javascript program to implement
// the above approach

// Print the required final sequence
function printSequence(A, n, k)
{
    
    // Stores the array element-index pairs
    var mp = new Map();

    // Stores the required sequence
    var sol = [];

    // Insert all array elements
    A.forEach(x => {
        sol.push(x);
    });

    for(var i = 0; i < n; i++) 
    {
        
        // If current element has
        // not occurred previously
        if (!mp.has(sol[i]) || 
             mp.get(sol[i]).size == 0) 
        {
            var tmp = new Set();
            tmp.add(i)
            mp.set(sol[i],tmp)
        }

        // Otherwise
        else
        {
            
            // Iterator to the first index
            // containing sol[i]
            var idxx = [...mp.get(sol[i])].sort(
                (a, b) => a - b)[0];

            var idx = idxx;

            // Remove that occurrence
            var x = mp.get(sol[i]);
            x.delete(idxx);
            mp.set(sol[i], x);

            // Increment by K
            sol[idx] += k;

            // Insert the incremented
            // element at that index
            if (!mp.has(sol[idx]))
                mp.set(sol[idx], new Set())
                
            x = mp.get(sol[idx]);
            x.add(idx);
            mp.set(sol[idx], x);

            x = mp.get(sol[i]);
            x.add(i);
            mp.set(sol[i], x);
        }
    }

    // Print the final sequence
    sol.forEach(x => {
        document.write(x + " ");
    });
}

// Driver Code
var N = 5;
var K = 6;
var A = [ 1, 4, 1, 1, 4 ];

printSequence(A, N, K);

// This code is contributed by importantly

</script> 

Output

7 10 7 1 4

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