Generate an array using given conditions from a given array

• Last Updated : 24 Nov, 2021

Given an array arr[] of size N, the task is to construct a new array B[] using elements of array A such that:

1. If the element is not present in B[], then append it to the end.
2. If the element is present in B[], then first increment its leftmost occurrence by 1 and then append this element to the end of the array.

Examples:

Input: arr[] = {1, 2, 1, 2}
Output: { 3, 2, 1, 2 }
Explanation:
arr = 1, B = {}. 1 is not present in B. So append it at the end. Therefore, B = {1}
arr = 2, B = {1}. 2 is not present in B. So append it at the end. Therefore, B[] = {1, 2}
arr = 1, B = {1, 2}. 1 is already present in B[]. So increment B by 1 and append 1 at the end. Therefore B[] = {2, 2, 1}
arr = 2, B = {2, 2, 1}. 2 is already present in B[]. So increment B by 1 and append 2 at the end. Therefore B[] = {3, 2, 1, 2}
Input: arr[] = {2, 5, 4, 2, 8, 4, 2}
Output: {3, 5, 5, 3, 8, 4, 2}

Naive Approach: For every element in the array A, check if it is present in array B or not. If the element exists, then increment the leftmost occurrence by one. Finally, add the element at the end of the array B[].
Time Complexity: O(N2)
Efficient Approach: The idea is to use a map to store all the indices of every element. The array B[] is generated as:

• For every element in the array arr[], it is checked if the element is already present in the array B[] or not.
• If the element is not present in the array B[], then the element is added to the array B[] and its index is added to the map. Since this is the first occurrence of the element, this index becomes the left-most index of the element.
• If the element is already present in the array B[], then the left-most index is returned and the element at that index is incremented by one. When the value is incremented, the left-most index of the old value is updated along with the index of the new value.

Below is the implementation of the above approach:

CPP

 // C++ program to generate an array// from a given array under the// given conditions #include using namespace std; // Function to generate an array// from a given array under the// given conditionsvoid newArray(int A[], int n){    // To maintain indexes of the    // elements in sorted order    unordered_map > idx;     // Initialize new array B    std::vector B;     // For every element in the given    // array arr[]    for (int i = 0; i < n; ++i) {         // Check if the element is present        // in the array B[]        if (idx.find(A[i]) != idx.end()) {             // Get the leftmost position            // in the array B            int pos = *idx[A[i]].begin();             // Remove the leftmost position            idx[A[i]].erase(idx[A[i]].begin());             // Increment the value at            // the leftmost position            B[pos]++;             // Insert new value position            // in the map            idx[B[pos]].insert(pos);        }         // Append arr[i] at the end        // of the array B        B.push_back(A[i]);         // Insert its position in hash-map        idx[A[i]].insert(i);    }     // Print the generated array    for (int i = 0; i < n; ++i)        cout << B[i] << " ";} // Driver codeint main(){    int arr[] = { 1, 2, 1, 2 };    int n = sizeof(arr) / sizeof(int);     newArray(arr, n);     return 0;}
Output:
3 2 1 2

Time Complexity: O(n)

Auxiliary Space: O(n)

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