How do you find the sum of a Submatrix?

How do you find the sum of a Submatrix?

The idea is to first create an auxiliary matrix aux[M][N] such that aux[i][j] stores sum of elements in submatrix from (0,0) to (i,j)….How to build aux[M][N]?

  1. Copy first row of mat[][] to aux[][]
  2. Do column wise sum of the matrix and store it.
  3. Do the row wise sum of updated matrix aux[][] in step 2.

How do you find the maximum sum of all Subarrays?

Simple Approach:

  1. Run a loop for i from 0 to n – 1, where n is the size of the array.
  2. Now, we will run a nested loop for j from i to n – 1 and add the value of the element at index j to a variable currentMax.
  3. Lastly, for every subarray, we will check if the currentMax is the maximum sum of all contiguous subarrays.

Which is true about kadane’s algorithm maximum?

Kadane’s algorithm is able to find the maximum sum of a contiguous subarray in an array with a runtime of O(n).

How do you find the submatrix of a matrix?

To find the submatrices it does this: take for example a 1×5 submatrix, what the code does is to fix the first line of the matrix and move step by step (along all the columns of the matrix) the submatrix from the left edge of the matrix to the right edge of the matrix, then the code fixes the second row of the matrix …

How do you find all submatrix of a matrix?

Let us suppose the index of an element be (X, Y) in 0 based indexing, then the number of submatrices (Sx, y) for this element will be in can be given by the formula Sx, y = (X + 1) * (Y + 1) * (N – X) * (N – Y).

How do you find the maximum sum of an array in Java?

  1. import java. util. Arrays; class Main.
  2. { // Function to find the maximum sum of a contiguous subarray. // in a given integer array.
  3. public static int kadane(int[] A) { // find the maximum element present in a given array.
  4. int max = Arrays. stream(A). max(). getAsInt();
  5. if (max < 0) { return max; }

Where is kadane’s algorithm used?

There are many applications of kadane’s algorithm and some of them are as mentioned below: Finding maximum subarray sum for a given array of integer. Used as an image processing algorithm. It can be used to solve the problems like “Station Travel in Order” and “Hotels Along the Coast”

Is kadane’s algorithm greedy?

Kadane’s Algorithm can be viewed both as a greedy and DP. As we can see that we are keeping a running sum of integers and when it becomes less than 0, we reset it to 0 (Greedy Part). This is because continuing with a negative sum is way more worse than restarting with a new range.

What is a principal submatrix?

A principal submatrix is a square submatrix obtained by removing certain rows and columns. The definition varies from author to author. According to some authors, a principal submatrix is a submatrix in which the set of row indices that remain is the same as the set of column indices that remain.

What is the maximum sum of a submatrix?

The maximum sum submatrix is [[7, 8, 3], [-2, -1, 2], [5, 5, 2], [2, 9, -5]] The maximum sum is 35 The time complexity of the proposed solution is O(N4)for an N × Nmatrix.

How to find the sum of all possible submatrices in matrices?

Naive Approach: The simplest approach is to generate all possible submatrices from the given matrix and calculate their sum. Finally, print the maximum sum obtained. Below is the implementation of the above approach:

How to get maxsum of a submatrix using Kadane’s algorithm?

After iterating all rows in the submatrix, perform Kadane’s algorithm on the array A [] and update the maximum sum maxSum as the maximum of maxSum and value obtained by performing the Kadane’s Algorithm in this step. After completing the above steps, print the value of maxSum as the result. Below is the implementation of the above approach:

How do you get the maximum sum of a subarray?

Initialize a variable, say maxSum as INT_MIN, to store the maximum subarray sum. Create a matrix prefMatrix [N] [M] that stores the prefix array sum of every row of the given matrix. If the value of i is 0, then set prefMatrix [i] [j] = A [i] [j].