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Puddle of Riddles!

reverse a linked list, preorder traversal without recursion, binary tree, recursion, string, array, searching, sorting, linked list, stack, queue, interview questions, graph, data structures.

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reverse a linked list, preorder traversal without recursion, binary tree, recursion, string, array, searching, sorting, linked list, stack, queue, interview questions, graph, data structures.
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Puddle of Riddles! | puddleofriddles.blogspot.com Reviews

https://puddleofriddles.blogspot.com

reverse a linked list, preorder traversal without recursion, binary tree, recursion, string, array, searching, sorting, linked list, stack, queue, interview questions, graph, data structures.

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puddleofriddles.blogspot.com puddleofriddles.blogspot.com
1

Puddle of Riddles!: October 2011

http://puddleofriddles.blogspot.com/2011_10_01_archive.html

Wednesday, October 26, 2011. Given an integer n, write code to calculate n 1, without using ,-, ,- ,*,/. Solution# 1: Start from the LSB and keep setting them to 0 till we find the first 0. And then set that 0 to 1. Here is the code,. Int increment(int n) { int i = 1; while(i&n) { n = n and i; i = i. Posted by Suchit Maindola. Labels: Bits n Bytes. Tuesday, October 25, 2011. Solution# 1: We can try all the permutations using the following code,. Int main() { int A; int B; int C; int count=0; for(A=0;A.

2

Puddle of Riddles!: November 2011

http://puddleofriddles.blogspot.com/2011_11_01_archive.html

Thursday, November 17, 2011. Program to find endianness. Solution #1: Finding endianness of a machine is to detect whether the machine is little endian or big endian. In a little endian machine the least significant byte is stored first whereas in a big endian machine the most significant byte is stored first. Here is the code,. Int main() { int i = 1; char* ptr = (char*)&i; if(*ptr) printf("Little Endian n"); else printf("Big Endian n"); return 0; }. Posted by Suchit Maindola. Med) { if( i. All subseque...

3

Puddle of Riddles!: Knapsack Problem

http://puddleofriddles.blogspot.com/2013/01/knapsack-problem.html

Monday, January 21, 2013. Solution: Below is the code using dynamic programming,. Define MAX(a,b) (a) (b)? W, choose V[n-1][w] */ if(w = weight) { V[n][w] = MAX( V[n-1][w], value V[n-1][w-weight] ); } else { V[n][w] = V[n-1][w]; } } } printf("Ans = %d n", V[N][W]); return 0; }. Posted by Suchit Maindola. Subscribe to: Post Comments (Atom).

4

Puddle of Riddles!: April 2012

http://puddleofriddles.blogspot.com/2012_04_01_archive.html

Friday, April 13, 2012. Determine direction of stack growth. Solution #1: The direction of stack growth depends on OS as well as on the architecture. Below is a program to detect the direction of stack growth,. Void b(int* x) { int y; / if address of y is less than address of x / then the direction is downwards, upwards otherwise if(&y. Posted by Suchit Maindola. Subscribe to: Posts (Atom). Determine direction of stack growth.

5

Puddle of Riddles!: June 2011

http://puddleofriddles.blogspot.com/2011_06_01_archive.html

Tuesday, June 28, 2011. Write a function to generate nth Fibonacci number. This can be done with both iteration and recursion. Solution #1: Here is an implementation using recursion,. Int fib rec(int n) { if(n = 0) return 0; if( n = 1) (n = 2) return 1; return fib rec(n-1) fib rec(n-2); }. Posted by Suchit Maindola. Wednesday, June 22, 2011. By adding arithmetic operators( , -, /, *) and parenthesis, make the following expression true: 3 1 3 6 = 8. Solution: ( 3 1) / 3)*6. Posted by Suchit Maindola.

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Puddle of Riddles!

Wednesday, January 23, 2013. Solution: Below is the code to solve a sudoku. N) { j = 0; i ; } /* * all rows are filled, print sol. and exit */ if(i N) { printMat(S); printf(" n"); exit(0); } /* * each cell can take a value belonging to [1,9] */ for(a=1; a. Posted by Suchit Maindola. Monday, January 21, 2013. Solution: Below is the code using dynamic programming,. Define MAX(a,b) (a) (b)? Posted by Suchit Maindola. Subscribe to: Posts (Atom).

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puddle of sun

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Not all who wander are lost | Slow down and trespass a little.

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Puddle on the Pitch: My Soccer Thoughts

Puddle on the Pitch: My Soccer Thoughts. Monday, May 12, 2014. Barclay's Premier League Best XI and Review 2014. Man City lifted the trophy for the second time in three seasons. It was all there this year. But now, let's get to some awards, highs, and lows. Let's start with a refresher on the Champions League Spots and teams that were relegated. BPL Champions League Qualification Spots-. Arsenal. (Playoff Qualification Round). Teams Relegated From the BPL-. Surprise Team of the Year: CRYSTAL PALACE.