Strange Docs

Unbounded Knapsack : Given a knapsack weight W and a set of N items with certain value Vi and weight W i , we need to calculate minimum amount that could make up this quantity exactly. To have a more clear understanding of this problem, see SPOJ-PIGBANK Sample Problem : We're given a box which can hold maximum W weight. There are N various coins each of value Vi and Weight Wi. We need to fill the box with total W weight with any number of coins of any value. Note that coins are unlimited.  100 ----- W 2 ------- N 1 1 ----- vi,wi 30 50 --- vi,wi Answer of the problem above is, 60. If we take 2 coins of value 30, weight 50, we'll get exactly weight 100 with a minimum value of 60. Algorithm : This is a DP solution. To get the solution for W=100, we'll solve subproblems first. Let dp[W+1]  be an array which will store the subproblems solutions. In the end, dp[W] will give us the minimum amount. Subproblems soultion will be the minimum of taking the coin Vk ...

Problem Statement: ANARC09A - Seinfeld Explanation: Minimum moves for balancing brackets

You’re given a non-empty string made in its entirety from opening and closing braces. Your task is to find the minimum number of “operations” needed to make the string stable. The definition of being stable is as follows: An empty string is stable. If S is stable, then {S} is also stable. If S and T are both stable, then ST (the concatenation of the two) is also stable. All of these strings are stable: {}, {}{}, and {{}{}}; But none of these: }{, {{}{, nor {}{. The only operation allowed on the string is to replace an opening brace with a closing brace, or visa-versa. Solution Heuristics: Traverse the string. When the number of opening braces( { ) matches the number of closing braces ( } ), the string is balanced. Else it's not balanced. So count both of them. Make a list. Insert only opening braces. If a closing brace is found delete one brace( if the size of the list is > 0 ) from the list. Because they would make a pair. If the list is empty then we'll conve...

Problem Statement: ENIGMATH - PLAY WITH MATH Heuristic: Ax - By = 0   has two possible cases:   a can divide b or if b can divide a  and you can figure out the rest.  a or b cannot divide the other so it will be A B - B A = 0, but we still have to find the minimum x & y.

Given an arbitrary series, we've to calculate the maximum sum confined to only one restriction- If we choose any number a [ i ] to calculate the sum, we can't choose a [ i+1 ] Let's take a series: a[ ]= {1,2,3,4,5,9} Linear Solution: Take two variable inc and exl . inc will store value if we choose a [ i ] to calculate the sum & exl will store value if we do not choose a [ i ], for any 0 ≤ i ≤ n. Now for every step, we have to calculate inc and exl . After iteration ends, max( inc,exl ) is the maximum non-contiguous sum. Roughly- Why inc = exl + a [ i ] ? because the already calculated exl is from i-2 . So if we choosed inc=  inc + a [ i ] , that would be a contiguous sum . exl is maximum of inc & exl right until   i-1 . Top Down solution( DP ): If you're familiar with dynamic programming, you know that, we need to calculate dp[n] - the maximum non contiguous sum at nth index. Now we can have two possible state, we'll select the maximum ...

Imagine a climber trying to climb on top of a wall.  A wall is constructed out of square blocks of equal size. From each block, the climber can reach three blocks of the row right above: one right on top, one to the right and one to the left ( unless right or left not available because that is end of the wall ). Find the least costly path from bottom to top where cost is the summation of costs of the blocks used on the path. Problem Representation: An NxM size grid where each cell has a positive cost C(i,j) associated with it. There exists three possible moves from a cell (i,j) - i+1,j-1     [ diagonally left ] i+1. j       [ above ] i+1.j+1     [ diagonally right ] Take an array dp[ n ][ m ] that will hold the cost for all cells. Initiate dp[ 0 ][ i ] = C(0,i)  for 0 ≤ i ≤ m . There are three cases to the recurrence: a cell might be in the middle (horizontally), on the leftmost or  on ...

Problem statement. The important part is ' Both players play optimally'; meaning: for every move a player can make a move that will lead closer to the winning move. So, when player one starts the game he can make sure the last move is his. Same goes for player two. and i know it is hard to believe.