Generate an array using given conditions from a given array
Given an array arr[] of size N, the task is to construct a new array B[] using elements of array A such that:
- If the element is not present in B[], then append it to the end.
- 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[0] = 1, B = {}. 1 is not present in B. So append it at the end. Therefore, B = {1}
arr[1] = 2, B = {1}. 2 is not present in B. So append it at the end. Therefore, B[] = {1, 2}
arr[2] = 1, B = {1, 2}. 1 is already present in B[]. So increment B[0] by 1 and append 1 at the end. Therefore B[] = {2, 2, 1}
arr[3] = 2, B = {2, 2, 1}. 2 is already present in B[]. So increment B[0] 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 <bits/stdc++.h> using namespace std; // Function to generate an array // from a given array under the // given conditions void newArray( int A[], int n) { // To maintain indexes of the // elements in sorted order unordered_map< int , set< int > > idx; // Initialize new array B std::vector< int > 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 code int main() { int arr[] = { 1, 2, 1, 2 }; int n = sizeof (arr) / sizeof ( int ); newArray(arr, n); return 0; } |
3 2 1 2
Time Complexity: O(n)
Auxiliary Space: O(n)