# Increasing Addition Codechef Solution

### We Are Discuss About CODECHEF SOLUTION

Increasing Addition Codechef Solution

## Increasing Addition Codechef Solution

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You have an array A of length N. In one operation, you can do the following:

• Select any subarray A_{L \ldots R} (1 \le L \le R \le N) and then add i to the i-th element of this subarray. Formally, we set A_j := A_j + (j – L + 1) for all L \le j \le R.

For example: if A = [2, 1, 5, 2] and we apply operation on A_{2 \ldots 4} then A becomes [2, 1 + 1, 5 + 2, 2 + 3] = [2, 2, 7, 5].

The minimum number of such operations required to sort A in non-decreasing order is termed as the goodness of A.

We want to process Q updates on A of the following form:

• \texttt{i x} : Set A_i := x where (1 \le i \le N).

Determine the goodness of A after each update.

### Input Format

• The first line contains a single integer T — the number of test cases. Then the test cases follow.
• The first line of each test case contains two integers N and Q — the size of the array A and the number of updates.
• The second line of each test case contains N space-separated integers A_1, A_2, \dots, A_N denoting the array A.
• The next Q lines each contain two integers i and x denoting the parameters of the update.

### Output Format

For each test case, output Q integers, the goodness of the array after each update.

### Constraints

• 1 \leq T \leq 10^4
• 2 \leq N \leq 10^5
• 1 \le Q \le N
• 1 \le A_i \le 10^9
• 1 \le x \le 10^9
• Sum of N over all test cases does not exceed 2 \cdot 10^5.

• Subtask 1 (50 points): 1 \le Q \le \min(5, N)
• Subtask 2 (50 points): Original constraints.

### Sample 1:

Input

Output

2
3 2
3 2 1
3 10
2 1
4 2
2 2 2 2
4 2
4 1

1
2
0
1


### Explanation:

Test Case 1: Initially A = [3, 2, 1].

• After first update: A = [3, 2, 10].
• Now we can apply the operation on A_{1 \ldots 2}. Then A becomes [4, 4, 10]. Therefore the goodness of A is 1.
• After second update: A = [3, 1, 10].
• Now we can apply the operation on A_{2 \ldots 2}A becomes [3, 2, 10]. Now again we can apply the operation A_{2 \ldots 2}A becomes [3, 3, 10]. Therefore the goodness of A is 2.

Test Case 2: Initially A = [2, 2, 2, 2].

• After first update: A = [2, 2, 2, 2].
• A is already in sorted order. Therefore the goodness of A is 0.
• After second update: A = [2, 2, 2, 1].
• Now we can apply the operation on A_{3 \ldots 4}A becomes [2, 2, 3, 3]. Therefore the goodness of A is 1.
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