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Brain Barfs | Mental musings and other arbitrary thoughts, rants or opinions about almost anything

Mental musings and other arbitrary thoughts, rants or opinions about almost anything

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Brain Barfs | Mental musings and other arbitrary thoughts, rants or opinions about almost anything | brainbarfs.wordpress.com Reviews

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About | Brain Barfs

https://brainbarfs.wordpress.com/about

Mental musings and other arbitrary thoughts, rants or opinions about almost anything. Leave a Reply Cancel reply. Enter your comment here. Fill in your details below or click an icon to log in:. Address never made public). You are commenting using your WordPress.com account. ( Log Out. You are commenting using your Twitter account. ( Log Out. You are commenting using your Facebook account. ( Log Out. You are commenting using your Google account. ( Log Out. Notify me of new comments via email.

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codinghangover.wordpress.com codinghangover.wordpress.com

Codeforces Beta Round #84 (Div. 1) A. Lucky Sum of Digits | coding hangover

https://codinghangover.wordpress.com/2015/03/20/codeforces-beta-round-84-div-1-a-lucky-sum-of-digits

Eat pray code :). Codeforces Beta Round #84 (Div. 1) A. Lucky Sum of Digits. March 20, 2015. Problem link: http:/ codeforces.com/problemset/problem/109/A. A minimum number must contain less number of digits as possible. So to find the minimum number whose digit sum is equal to given n, we have to look into the case 1 and if it fails we have to move to case 2.If case 2 also fails then such minimum lucky number cannot be formed. Case 1:if the number can be formed using digits 7 alone. Follow us on Facebook!

codinghangover.wordpress.com codinghangover.wordpress.com

vishnujayvel | coding hangover

https://codinghangover.wordpress.com/author/vishnujayvel

Eat pray code :). Spoj( ONEZERO ) – Ones and zeros. May 1, 2015. Problem link: http:/ www.spoj.com/problems/ONEZERO/. Suppose the number that you want is X. X mod N = 0. So you need to store only N states i.e. 0 to n-1. Start with 1. Implement bfs approach. If the current state modulo is Y, append 0 to it i.e calculate Y*10 0 and find its modulo which will lead you to a new modulo state. Similary append 1 to Y. i.e calculate Y*10 1 and find its modulo. Consisting only of digits 1 and 0 beginning with 1).

codinghangover.wordpress.com codinghangover.wordpress.com

Spoj( ONEZERO ) – Ones and zeros | coding hangover

https://codinghangover.wordpress.com/2015/05/01/spoj-onezero-ones-and-zeros

Eat pray code :). Spoj( ONEZERO ) – Ones and zeros. May 1, 2015. Problem link: http:/ www.spoj.com/problems/ONEZERO/. Suppose the number that you want is X. X mod N = 0. So you need to store only N states i.e. 0 to n-1. Start with 1. Implement bfs approach. If the current state modulo is Y, append 0 to it i.e calculate Y*10 0 and find its modulo which will lead you to a new modulo state. Similary append 1 to Y. i.e calculate Y*10 1 and find its modulo. Consisting only of digits 1 and 0 beginning with 1).

codinghangover.wordpress.com codinghangover.wordpress.com

Codeforces #264 (Div. 2) C. Gargari and Bishops | coding hangover

https://codinghangover.wordpress.com/2015/03/18/codeforces-264-div-2-c-gargari-and-bishops

Eat pray code :). Codeforces #264 (Div. 2) C. Gargari and Bishops. March 18, 2015. Problem link: http:/ codeforces.com/problemset/problem/463/C. To find 2 positions such that bishops in these positions cannot attach each other, we have to select 2 coordinates such that sum of x,y(row,column) in one coordinate is even and odd in other one. Codeforces #296 (Div. 2) B. Error Correct System. Codeforces Beta Round #84 (Div. 1) A. Lucky Sum of Digits →. Leave a Reply Cancel reply. Enter your comment here.

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Solving K-Palindrome problem using Edit distance algorithm | coding hangover

https://codinghangover.wordpress.com/2015/03/22/solving-k-palindrome-problem-using-edit-distance-algorithm

Eat pray code :). Solving K-Palindrome problem using Edit distance algorithm. March 22, 2015. A k-palindrome is a string which transforms into a palindrome on removing at most k characters. Given a string S, and an interger K, print “YES” if S is a k-palindrome; otherwise print “NO”. S has at most 20,000 characters. Input – abxa 1. Output – YES. Input – abdxa 1. Output – No. A string S is K-palindrome if editDistance(S,reverse(S) =2*K. This will be more clear with an example. Let S=madtam and K=1. Now th...

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Codeforces #224 (Div. 2) C. Arithmetic Progression | coding hangover

https://codinghangover.wordpress.com/2015/03/21/codeforces-224-div-2-c-arithmetic-progression

Eat pray code :). Codeforces #224 (Div. 2) C. Arithmetic Progression. March 21, 2015. Problem link – http:/ codeforces.com/problemset/problem/382/C. Codeforces Beta Round #84 (Div. 1) A. Lucky Sum of Digits. Solving K-Palindrome problem using Edit distance algorithm →. Leave a Reply Cancel reply. Enter your comment here. Fill in your details below or click an icon to log in:. Address never made public). You are commenting using your WordPress.com account. ( Log Out. Notify me of new comments via email.

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Spoj (QUE1) Queue (Rookie) | coding hangover

https://codinghangover.wordpress.com/2015/03/25/spoj-que1-queue-rookie

Eat pray code :). Spoj (QUE1) Queue (Rookie). March 25, 2015. Problem link – http:/ www.spoj.com/problems/QUE1/. Vector of pairs is used to store the height and count of taller person standing before. This problem can be solved by sorting the vector based on height. Assign the final position for each person i in the answer array by moving them C empty positions from the index 0 where. C = count of taller person who are standing before person i. Detailed explanation is in the comments of the code below.

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Brain Barfs | Mental musings and other arbitrary thoughts, rants or opinions about almost anything

Mental musings and other arbitrary thoughts, rants or opinions about almost anything. Apologies, but no results were found for the requested archive. Perhaps searching will help find a related post. Follow Blog via Email. Enter your email address to follow this blog and receive notifications of new posts by email. Join 1,760 other followers. Follow me on Twitter. Create a free website or blog at WordPress.com.

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