SDE - Intern

You are given a binary matrix with ‘M’ rows and ‘N’ columns initially consisting of all 0s. 'Q' queries follow. The queries can be of 4 types:

Query 1: 1 R index
Query 2: 1 C index 
Query 3: 2 R index
Query 4: 2 C index

In each query, the first input is the type of the query, the second input is whether we have to consider the row ('R') or the column ('C') and the third input is the index of the row/column. 

For each type 1 query, we need to flip the elements of the row/column having the given index. 

For each type 2 query, we have to output the number of zeros present in the row/column having the given index. 

Note:

Note that the matrix is a binary matrix, meaning that it only contains either 0 or 1.

Example :

Given M = 3, N = 3, 
Queries : 1R1, 1R2, 2C1

So, in the above example the change in the matrix would look like this:

Next query 2C1 will return the count of the number of zeroes in the 1st column: 1

You have been given an array 'ARR' of ‘N’ integers. You have to return the minimum number of jumps needed to reach the last index of the array i.e ‘N - 1’.

From index ‘i’, we can jump to an index ‘i + k’ such that 1<= ‘k’ <= ARR[i] .

'ARR[i]' represents the maximum distance you can jump from the current index.

If it is not possible to reach the last index, return -1.

Note:

Consider 0-based indexing.

Example:

Consider the array 1, 2, 3, 4, 5, 6 
We can Jump from index 0 to index 1
Then we jump from index 1 to index 2
Then finally make a jump of 3 to reach index N-1

There is also another path where
We can Jump from index 0 to index 1
Then we jump from index 1 to index 3
Then finally make a jump of 2 to reach index N-1

So multiple paths may exist but we need to return the minimum number of jumps in a path to end which here is 3.

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