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

• Difficulty Level : Hard
• Last Updated : 02 Jun, 2021

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[] = 

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 using namespace std; // Print the required final sequencevoid printSequence(vector& A,                   int n, int k){    // Stores the array element-index pairs    unordered_map > mp;     // Stores the required sequence    vector 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 Codeint main(){    int N = 5;    int K = 6;    vector A = { 1, 4, 1, 1, 4 };    printSequence(A, N, K);}

Javascript


Output:
7 10 7 1 4

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

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