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imperfect_notes/pragy/Maths 3 - Modular Arithmetic.md

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+Maths 3 - Modular Arithmetic
+----------------------------
+
+
+
+Arithmetic of Remainders
+
+$x % m$ gives a value $\in [0, m-1]$
+
+
+Say m = 4
+
+**Property 1 - Modulus repeats**
+![97402b22.png](/users/pragyagarwal/Boostnote/attachments/b40d7037-2666-4b31-b623-936a12cc4244/b445a06d.png)
+
+Since 1, 5, 10 have the same modulus, they can be said to be "modular equivalent wrt 5"
+
+$0 \equiv 5 \equiv 10 \mod 5$
+$1 \equiv 6 \equiv 11 \mod 5$
+-- --
+
+Linearity of remainders
+-----------------------
+
+Remainder of sums is the sum of remainders (modular-sum)
+
+**Incorrect:** $(a+b) \% m = a \% m + b \% m$
+
+This is wrong, because (m-1) + (m-1) = (2m-2) can be >= m
+
+**Correct:** $(a+b) \% m = \Bigl ( a \% m + b \% m \Bigr ) \% m$
+
+**Linearity works for both +ve and negative.**
+But, we need numbers to be $\in [0, m-1]$
+
+So, negative modulus values can be fixed by simply adding $m$
+
+$(10 - 4) \% 7$
+$= (10 \% 7 - 4 \% 7) \% 7$
+$= (3 - 4) \% 7$
+$= -1 \% 7$
+$= -1 + 7$
+$= 6$
+
+**Distributive over multiplication, addition, substraction but not division.**
+
+-- --
+
+Count Pairs
+-----------
+
+
+> Given A[N]. non-negative integers.
+> Given m
+> Count number of pairs (i, j), such that a[i] + a[j] is divisible by m
+> 
+
+
+**Brute Force:** $O(n^2)$
+
+Example:
+> A = [2 2 1 7 5 3], m = 4
+> answer = 5
+> (2,2), (1, 7), (1, 3), (7, 5), (5, 3)
+
+**Optimized:**
+Instead of thinking of remainders of sum, think of sum of remainders.
+
+That is, if I've an element which gives remainder $x$, then it can be paired with an element which gives remainder $m-x$
+![5dd03bec.png](/users/pragyagarwal/Boostnote/attachments/b40d7037-2666-4b31-b623-936a12cc4244/bf41b9e2.png =300x)
+Keep counts for each remainder. Finally, count the number of pairs
+
+> 0 - 0
+> 1 - 2
+> 2 - 2
+> 3 - 2
+> So, $\dfrac{c_0 (c_0-1)}2 + (c_1 \times c_3) + \dfrac{c_2 (c_2-1)}2$
+> $= 0 + 4 + 2 = 6$
+> 
+
+Special cases for remainders is 0, n/2
+Note: loop till m/2 so that you don't double count
+
+```python
+total = count[0] choose 2
+if m % 2 ==0:
+    total += count[m/2] choose 2
+else:
+    total += count[m//2] + count[m//2 + 1]
+
+for i <- 1 .. m/2-1:
+    total += count[i] * count[m-i]
+```
+
+
+Complexity: Time: $O(n + m)$ Space: $O(m)$
+-- --
+
+Count Triplets
+--------------
+
+> Given A[N] with non-negative integers
+> Given m, find the number of triplets such that sum is divisible by m
+> 
+
+Think in terms of the last question
+
+**Brute Force**: $O(n^3)$
+
+**With last approach:** $O(n + m^2)$
+
+1. Bucket
+2. Choose 2 mods: a, b. Third mod c = m - (a+b)
+3. Count total
+4. All 3 from same bucket - for 0 and m//3
+5. 2 from same bucket, one from other bucket
+6. all 3 from different buckets
+
+![bb5890c2.png](/users/pragyagarwal/Boostnote/attachments/b40d7037-2666-4b31-b623-936a12cc4244/bb5890c2.png =400x)
+
+So, fix two buckets, so 3rd can be chosen
+
+Enforce two things:
+1. i <= j <= k   otherwise, repetitions can happen
+2. i + j + k = 0 or i + j + k = m
+
+
+```python
+total = 0
+
+for i <- 0 .. m-1:
+    for j <- i .. m-1:
+        k = (m - i + j) % m # explain the case when i, j are 0
+        # explain that this mod should handle -ve
+        # python. for c++/java, do it yourself
+        if i == j == k:
+            total += count[i] choose 3
+        elif i == j:
+            total += count[i] choose 2 * count[k]
+        elif i == k:
+            total += count[i] choose 2 * count[j]
+        elif j == k:
+            total += count[j] choose 2 * count[i]
+        else:
+            total += count[i] * count [j] * count[k]
+
+```
+
+Time: $O(n + m^2)$, Space: $O(m)$
+
+
+-- --
+
+Balanced Paranthesis Count
+--------------------------
+
+> Given N pairs of paranthesis
+> Find number of distinct balanced paranthesis
+> 
+> meaning of balanced: for every prefix, number of opening braces should be >= number of closing braces
+> 
+
+Example
+> 0 pairs
+> 0
+> 
+> 1 pair: ()
+> 1
+> 
+> 2 pairs: ()(), (())
+> 2
+> 
+> 3 pairs: ()()(), ()(()), (())(), (()()), ((()))
+> 5
+
+Say we want to construct for n = 3
+Choose how many pairs must go inside
+
+![33f72a14.png](/users/pragyagarwal/Boostnote/attachments/b40d7037-2666-4b31-b623-936a12cc4244/33f72a14.png =400x)
+
+![962dd530.png](/users/pragyagarwal/Boostnote/attachments/b40d7037-2666-4b31-b623-936a12cc4244/962dd530.png =300x)
+
+
+$C_0 = 1$
+
+$C_n = C_0 C_{n-1} + C_1 C_{n-2} + \cdots C_{n-1}C_0$
+
+$\displaystyle C_n = \sum\limits_{i=0}^{n-1} C_i C_{n-i-1}$
+
+-- --
+
+Catalan numbers
+---------------
+
+
+Same as above
+
+Time to find nth catlan number = $O(n^2)$ (because we need to compute numbers before it)
+
+-- --
+
+
+Number of Paths
+---------------
+
+> Given $m \times n$ grid, find the number of paths from (0,0) to (m-1, n-1)
+> Can only move right and down.
+
+
+Approach:
+- Every path will have exactly (n-1) right steps, and (m-1) down steps.
+- Only the order differs
+
+So, $\dfrac{(n+m-2)!}{(n-1)! \cdot (m-1)!}$
+
+Complexity? O(n+m)
+
+-- --
+
+Number of lower triangle paths in Square Matrix
+-----------------------------------------------
+
+
+> Given $n \times n$ (square) grid, find the number of paths from (0,0) to (m-1, n-1)
+> Can only move right and down.
+> Exists only in teh lower triangle
+> 
+
+- First move must be down
+- Number of Rs < Number of Ds at every step
+
+Same as balanced parans
+![7c611269.png](/users/pragyagarwal/Boostnote/attachments/b40d7037-2666-4b31-b623-936a12cc4244/7c611269.png)
+
+$C_{n-1}$, because n-1 pairs of moves for $n \times n$ matrix
+