Add all my notes

imperfect_notes/pragy/Arrays + 2d Arrays.md

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+Arrays + 2d Arrays
+------------------
+
+anish - disruptor
+
+
+Vivek - May EliteX
+
+-- --
+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
+> 
+
+![10417df8.png](/users/pragyagarwal/Boostnote/attachments/58e5d3fa-9fb3-4fac-8589-df43c99e4851/10417df8.png =250x)
+
+
+**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: $O(n^6)$
+
+```python
+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**
+
+![50177743.png](/users/pragyagarwal/Boostnote/attachments/58e5d3fa-9fb3-4fac-8589-df43c99e4851/50177743.png =150x)
+
+- 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
+
+![f9420692.png](/users/pragyagarwal/Boostnote/attachments/58e5d3fa-9fb3-4fac-8589-df43c99e4851/f9420692.png =400x)
+
+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 = $a[i][j] \times (i+1)(j+1)(n-i)(n-j)$
+
+
+
+$O(n^2)$
+
+-- --
+
+This is called reverse lookup
+(not to be confused with inverting the problem)
+
+-- --
+
+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
+
+![b3741590.png](/users/pragyagarwal/Boostnote/attachments/58e5d3fa-9fb3-4fac-8589-df43c99e4851/b3741590.png)
+
+First by rows, then by columns (or vice versa)
+
+Now, add and substract sum to get the desired sums
+![89352278.png](/users/pragyagarwal/Boostnote/attachments/58e5d3fa-9fb3-4fac-8589-df43c99e4851/89352278.png)
+
+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**
+
+-- --
+
+
+Max Submatrix Sum
+------------------
+
+> Given Matrix
+> All rows are sorted
+> All columns are sorted
+> Find the max sum submatrix
+> 
+
+![cf7e8a03.png](/users/pragyagarwal/Boostnote/attachments/58e5d3fa-9fb3-4fac-8589-df43c99e4851/cf7e8a03.png =300x)
+
+**Brute force** $O(n^6)$
+**All submatrices using prefix sum** $O(n^4)$
+
+**Optimal**
+Create suffix sums of matrix
+find max value
+
+$O(n^2)$
+
+
+-- --
+
+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**
+![237e989f.png](/users/pragyagarwal/Boostnote/attachments/58e5d3fa-9fb3-4fac-8589-df43c99e4851/237e989f.png =300x)
+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: $O(m+n)$
+
+-- --
+
+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
+
+
+
+
+--- ---
+
+--- ---
+
+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