11. Adding Ones

The problem can be found at the following link: Question Link

My Approach

To solve the problem, I have used the following approach:

• Initialize a vector out of size n+1 with all elements initially set to 0. This vector will be used to store the counts of each number.

• Traverse the updates array and increment the count in the out vector for each number encountered.

• Initialize the first element of the arr array as the count of 1 in the out vector.

• Traverse the arr array starting from the second element. Set each element as the sum of the previous element in the arr array and the count of the corresponding number in the out vector.

• Finally, the arr array stores the cumulative counts of each number from 1 to n.

Time and Auxiliary Space Complexity

• Time Complexity: O(k + n), where k is the size of the updates array and n is the size of the arr array. We traverse the updates array once and the arr array once, performing constant-time operations in each iteration.

• Auxiliary Space Complexity: O(n) since we use an additional vector out of size n+1 to store the counts of each number.

Code (C++)

class Solution{
    public:
    void update(int arr[], int n, int updates[], int k){
        vector<int> out(n+1, 0);
        for(int i = 0; i < k; ++i){
            ++out[updates[i]];
        }
        arr[0] = out[1];
        for(int i = 2; i <= n; ++i){
            arr[i-1] = arr[i-2] + out[i];
        }
    }
};

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