Lecture_Notes/imperfect_notes/pragy/Arrays + 2d Arrays.md

Arrays + 2d Arrays

anish - disruptor

Vivek - May EliteX

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pragy@interviewbit.com

Sum of all Submatrices

Given matrix M_{n \times n}, find the sum of all submatrices of M

Example:

1 1 1 1

sum = 1 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 4 = 16


Brute Force Go over each submatrix, and add

• Submatrix is a rectangle - topleft and bottomright points
• So, four for loops to decide a submatrix (two for topleft, 2 for bottomright)
• then, two for loops to go over each element in the submatrix

Complexity: O(n^6)

total = 0

top <- 0 .. n
    left <- 0 .. n
        bottom <- top .. n
            right <- left .. n
                i <- top .. bottom
                    j <- left .. right
                        total += M[i][j]

Optimized


• Element i, j appears in multiple submatrices
• If a[i][j] occurs in k submatrices, its contribution to the total is k \times a[i][j]
• So, we need to count the number of occurances of each i, j


Just count the number of submatrices that contain i, j - choose any top right from allowed, choose any bottom right from allowed

For top-right: (i+1)(j+1) For bottom-left: (n-i) (n-j)

Total for i, j = (i+1)(j+1)(n-i)(n-j)

So, contribution = a[i][j] \times (i+1)(j+1)(n-i)(n-j)

O(n^2)

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This is called reverse lookup (not to be confused with inverting the problem)

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Any Submatrix Sum

Given M_{m \times n} q queries, q is large, of the form (top-left, bottom-right) Find the sum of the submatrix specified by each query

Naive

O(qmn)

Optimized Calculate Prefix Sums

• explain for 1d array
• extend to 2d array

First by rows, then by columns (or vice versa)

Now, add and substract sum to get the desired sums

Prefix Sum matrix = P

sum(top, left, bottom, right) = P(bottom, right) - P(bottom, left - 1) - P(top-1, right) + P(top-1, left-1)

Corner cases

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Max Submatrix Sum

Given Matrix All rows are sorted All columns are sorted Find the max sum submatrix


Brute force O(n^6) All submatrices using prefix sum O(n^4)

Optimal Create suffix sums of matrix find max value

O(n^2)

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Find number ZigZag

Given matrix Each row and column is sorted Find a given number k

Approach 1

Binary search each row/column

O(n \log n)

Optimal  Start from top right of the remaining matrix

If k is larger, scrape off the row If k is smaller, scrape off the column

In every iteration, we scrape off one row or column

Complexity: O(m+n)

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Chunks

Given Array A[N] A[i] \in \{0 .. N-1\} All elements are distinct

The array is a permutation

1. Split the array into chunks C_1, C_2, C_3 \ldots
2. Sort individual chunks
3. Concatenate

The array after these operatio0ns should be sorted Find the max number of chunks that the array can be split into

minimizing? ans: 1

sorted array (max)? ans: |A| reverse sorted? 1

Observation

For i to j to be a valid chunk, it has to contain all the numbers i to j

Approach

Compute max from left to right. If at any point, max == i, we have a chunk So, do a chunk++

If max > i, can we jump to i? No. because there might be something even greater on the way Can max < i? No

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Lecture bookmarks

mine - 00:04:00 Lecture Start 00:04:29 Q1 - Find sum of all submatrices of given Matrix 00:05:28 Q1 - Example 00:08:00 Q1 - Brute Force Approach 00:15:30 Q1 - Optimized (Intuition) 00:19:28 Q1 - Optimized (Approach) 00:24:56 Q1 - Optimized - Formula re-explanation 00:27:30 Pattern: Reverse lookup 00:28:16 Q2 - Answer many submatrix sum Queries 00:30:30 Q2 - Brute Force Approach 00:31:20 Q2 - 1D Array Prefix Sum 00:35:20 Q2 - Extending prefix-sum to Matrices 00:46:25 Q2 - Corner Cases 00:48:07 Q3 - Given Matrix with rows & columns sorted, find Largest sum submatrix 00:49:42 Q3 - Brute Force Approach 00:50:47 Q3 - Optimization 1 00:52:38 Q3 - Optimization 2 01:00:07 Q4 - Given Matrix with rows & columns sorted, search for some key k 01:00:47 Q4 - Binary Search Approach 01:03:27 Q4 - ZigZag approach 01:15:22 Q5 - Max number of chunks 01:18:18 Q5 - Example 01:20:18 Q5 - Special Cases 01:22:47 Q5 - Observations 01:25:20 Q5 - Optimized 01:29:15 Q5 - Example Run 01:30:15 Q5 - Pseudo Code 01:32:20 Doubt Clearing