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.

**Examples:**

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,

`// C++ program to print weights for alternating ` `// the weighting scale ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// DFS method to traverse among states of weighting scales ` `bool` `dfs(` `int` `residue, ` `int` `curStep, ` `int` `wt[], ` `int` `arr[], ` ` ` `int` `N, ` `int` `steps) ` `{ ` ` ` `// If we reach to more than required steps, ` ` ` `// return true ` ` ` `if` `(curStep > steps) ` ` ` `return` `true` `; ` ` ` ` ` `// Try all possible weights and choose one which ` ` ` `// returns 1 afterwards ` ` ` `for` `(` `int` `i = 0; i < N; i++) ` ` ` `{ ` ` ` `/* Try this weight only if it is greater than ` ` ` `current residueand not same as previous chosen ` ` ` `weight */` ` ` `if` `(arr[i] > residue && arr[i] != wt[curStep - 1]) ` ` ` `{ ` ` ` `// assign this weight to array and recur for ` ` ` `// next state ` ` ` `wt[curStep] = arr[i]; ` ` ` `if` `(dfs(arr[i] - residue, curStep + 1, wt, ` ` ` `arr, N, steps)) ` ` ` `return` `true` `; ` ` ` `} ` ` ` `} ` ` ` ` ` `// if any weight is not possible, return false ` ` ` `return` `false` `; ` `} ` ` ` `// method prints weights for alternating scale and if ` `// not possible prints 'not possible' ` `void` `printWeightsOnScale(` `int` `arr[], ` `int` `N, ` `int` `steps) ` `{ ` ` ` `int` `wt[steps]; ` ` ` ` ` `// call dfs with current residue as 0 and current ` ` ` `// steps as 0 ` ` ` `if` `(dfs(0, 0, wt, arr, N, steps)) ` ` ` `{ ` ` ` `for` `(` `int` `i = 0; i < steps; i++) ` ` ` `cout << wt[i] << ` `" "` `; ` ` ` `cout << endl; ` ` ` `} ` ` ` `else` ` ` `cout << ` `"Not possible\n"` `; ` `} ` ` ` `// Driver code to test above methods ` `int` `main() ` `{ ` ` ` `int` `arr[] = {2, 3, 5, 6}; ` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(` `int` `); ` ` ` ` ` `int` `steps = 10; ` ` ` `printWeightsOnScale(arr, N, steps); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

Output:

2 3 2 3 5 6 5 3 2 3

This article is contributed by **Utkarsh Trivedi**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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