# 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:

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[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}

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

 `// 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 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; ` `} `

Output:

```3 2 1 2
```

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