SDE - 1

You have been given a non-empty grid ‘mat’ with 'n' rows and 'm' columns consisting of only 0s and 1s. All the rows are sorted in ascending order.

Your task is to find the index of the row with the maximum number of ones.

Note: If two rows have the same number of ones, consider the one with a smaller index. If there's no row with at least 1 zero, return -1.

Example:

Input: 'n' = 3, 'm' = 3, 'mat' = [[1, 1, 1], [0, 0, 1], [0, 0, 0]]

Output: 0

Explanation: The row with the maximum number of ones is 0 (0 - indexed).

You are given an array “ARR” of 'N' integers and an integer 'K'. You can move from any index 'i' to index 'j' if j ≤ i + K. The cost of moving from one index 'i' to the other index 'j' is abs(ARR[j] – ARR[i]). Your task is to find the minimum cost to reach the end of the array from the beginning of the array when a maximum jump of 'K' is allowed.

For example:

If the given array is [10, 3, 40, 5, 25] and K is 2 then the minimum cost would be 29. 
Since K = 2, the optimal way to reach the end of the array with minimum cost is to take a jump to 1st index from 0th index with the cost of abs(3 - 10)  i.e 7 and then we take a jump of 2 from 1st index to the 3rd index with the cost of abs(5 - 3). i.e 2. Then we take a jump of 1 from 3rd index to the last index with the cost of abs(25 - 5) .ie 20. 
Therefore the minimum cost to reach the end of the array is (7 + 2 + 20) i.e 29.

You are given an array “ARR” of 'N' integers and an integer 'K'. You can move from any index 'i' to index 'j' if j ≤ i + K. The cost of moving from one index 'i' to the other index 'j' is abs(ARR[j] – ARR[i]). Your task is to find the minimum cost to reach the end of the array from the beginning of the array when a maximum jump of 'K' is allowed.

For example:

If the given array is [10, 3, 40, 5, 25] and K is 2 then the minimum cost would be 29. 
Since K = 2, the optimal way to reach the end of the array with minimum cost is to take a jump to 1st index from 0th index with the cost of abs(3 - 10)  i.e 7 and then we take a jump of 2 from 1st index to the 3rd index with the cost of abs(5 - 3). i.e 2. Then we take a jump of 1 from 3rd index to the last index with the cost of abs(25 - 5) .ie 20. 
Therefore the minimum cost to reach the end of the array is (7 + 2 + 20) i.e 29.