Given a weighting scale and an array of different positive weights where we have an infinite supply of each weight. Our task is to put weights on left and right pans of scale one by one in such a way that pans move to that side where weight is put i.e. each time, pans of scale moves to alternate sides.
- We are given another integer ‘steps’, times which we need to perform this operation.
- Another constraint is that we can’t put same weight consecutively i.e. if weight w is taken then in next step while putting the weight on opposite pan we can’t take w again.
Let weight array is [7, 11] and steps = 3 then 7, 11, 7 is the sequence in which weights should be kept in order to move scale alternatively. Let another weight array is [2, 3, 5, 6] and steps = 10 then, 3, 2, 3, 5, 6, 5, 3, 2, 3 is the sequence in which weights should be kept in order to move scale alternatively.
This problem can be solved by doing DFS among scale states.
- We traverse among various DFS states for the solution where each DFS state will correspond to actual difference value between left and right pans and current step count.
- Instead of storing weights of both pans, we just store the difference residue value and each time chosen weight value should be greater than this difference and shouldn’t be equal to previously chosen value of weight.
- If it is, then we call the DFS method recursively with new difference value and one more step.
Please see below code for better understanding,
2 3 2 3 5 6 5 3 2 3
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- Count the nodes of a tree whose weighted string is an anagram of the given string
- Count the nodes of a tree whose weighted string does not contain any duplicate characters
- Print the degree of every node from the given Prufer sequence
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- Count the nodes of the given tree whose weight has X as a factor
- Count the nodes of the given tree whose weighted string is a palindrome
Improved By : PranchalKatiyar