🚀 08-Aug-2020

Author: sureshmangs

Diff for: competitive programming/leetcode/171. Excel Sheet Column Number.cpp

@@ -0,0 +1,51 @@
+Given a column title as appear in an Excel sheet, return its corresponding column number.
+
+For example:
+
+    A -> 1
+    B -> 2
+    C -> 3
+    ...
+    Z -> 26
+    AA -> 27
+    AB -> 28
+    ...
+Example 1:
+
+Input: "A"
+Output: 1
+Example 2:
+
+Input: "AB"
+Output: 28
+Example 3:
+
+Input: "ZY"
+Output: 701
+
+
+Constraints:
+
+1 <= s.length <= 7
+s consists only of uppercase English letters.
+s is between "A" and "FXSHRXW".
+
+
+
+
+
+
+
+
+class Solution {
+public:
+    int titleToNumber(string s) {
+        int res=0, k=0;
+        int n=s.length();
+        for(int i=n-1;i>=0;i--){
+            int ch=(s[i]-'A')+1;
+            res+=ch*pow(26,k++);
+        }
+        return res;
+    }
+};

Diff for: mathematical/catalan.cpp

@@ -1,3 +1,47 @@
+/*
+Given a number N. The task is to find the nth Catalan number.
+The first few Catalan numbers for n = 0, 1, 2, 3, … are 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, …
+
+Input:
+The first line of input contains a single integer T which denotes the number of test cases. The first line of each test case contains a single integer N.
+
+Output:
+For each test case, in a new line print the Catalan number at position N.
+Note: Positions start from 0 as shown above.
+
+Expected Time Complexity: O(N).
+Expected Auxiliary Space: O(N).
+
+Constraints:
+1 <= T <= 100
+1 <= N <= 100
+
+Example:
+Input:
+3
+5
+4
+10
+
+Output:
+42
+14
+16796
+Explanation:
+From the given first few Catalan's number we can easily the answer for first two test cases.
+*/
+
+
+
+
+
+
+
+
+
+
+
+
 #include<bits/stdc++.h>
 #include <boost/multiprecision/cpp_int.hpp>
 using namespace boost::multiprecision;

Diff for: mathematical/check_if_prime.cpp

@@ -50,3 +50,18 @@ int main(){
 
 return 0;
 }
+
+
+
+
+/*
+Prime number is in form of 6p+1, 6p-1
+example
+6p divisible by 6
+6p+1
+6p+2 divisible by 2
+6p+3 divisible by 3
+6p+4 divisible by 4
+6p+5 = 6p+6-1 =  6(p+1)-1 = 6X-1
+6p+6 divisible by 6
+*/

Diff for: mathematical/extended_eucledian_algorithm.cpp

@@ -1,3 +1,35 @@
+/*
+As seen above, x and y are results for inputs a and b,
+   a.x + b.y = gcd                      ----(1)
+
+And x1 and y1 are results for inputs b%a and a
+   (b%a).x1 + a.y1 = gcd
+
+When we put b%a = (b - (⌊b/a⌋).a) in above,
+we get following. Note that ⌊b/a⌋ is floor(b/a)
+
+   (b - (⌊b/a⌋).a).x1 + a.y1  = gcd
+
+Above equation can also be written as below
+   b.x1 + a.(y1 - (⌊b/a⌋).x1) = gcd      ---(2)
+
+After comparing coefficients of 'a' and 'b' in (1) and
+(2), we get following
+   x = y1 - ⌊b/a⌋ * x1
+   y = x1
+
+*/
+
+
+
+
+
+
+
+
+
+
+
 #include<bits/stdc++.h>
 using namespace std;
 

Diff for: mathematical/gcd_a_b.cpp

@@ -20,3 +20,9 @@ int main(){
     }
 return 0;
 }
+
+
+
+
+
+// TC : O(Log min(a, b))

Diff for: mathematical/x_power_n_negative_decimal.cpp

@@ -1,3 +1,27 @@
+/*
+Implement pow(x, n), which calculates x raised to the power n (xn).
+
+Example 1:
+
+Input: 2.00000, 10
+Output: 1024.00000
+Example 2:
+
+Input: 2.10000, 3
+Output: 9.26100
+Example 3:
+
+Input: 2.00000, -2
+Output: 0.25000
+Explanation: 2-2 = 1/22 = 1/4 = 0.25
+*/
+
+
+
+
+
+
+
 #include<bits/stdc++.h>
 using namespace std;
 

Diff for: mathematical/x_power_y.cpp

@@ -1,3 +1,16 @@
+/*
+Input : x = 2, n = 3
+Output : 8
+
+Input : x = 7, n = 2
+Output : 49
+*/
+
+
+
+
+
+
 #include<bits/stdc++.h>
 using namespace std;
 

Diff for: matrix/rotate_matrix_90_anticlockwise.cpp

@@ -90,3 +90,23 @@ int main(){
 
 return 0;
 }
+
+
+
+
+/*
+Matrix rotation by 90 degrees can be done in
+
+clockwise
+anticlockwise
+
+To rotate a matrix clockwise by 90 degrees:
+
+Find the transpose of the matrix.
+Reverse every rows of the matrix.
+
+To rotate a matrix anticlockwise by 90 degrees,
+
+Find the transpose of the matrix.
+Reverse every columns of the matrix.
+*/

Diff for: string/excel_sheet_column_number.cpp

@@ -66,3 +66,20 @@ int main(){
 
 return 0;
 }
+
+
+
+
+
+
+/* Reverse of above question
+int titleToNumber(string s) {
+        int res=0, k=0;
+        int n=s.length();
+        for(int i=n-1;i>=0;i--){
+            int ch=(s[i]-'A')+1;
+            res+=ch*pow(26,k++);
+        }
+        return res;
+    }
+    *.