## Determine winner of game by converting two consecutive 1s to 0s in C++## Introduction to the Problem:The problem description revolves round a very simple game by the provided moves. After that, the player who has a successful combination of moves will have a chance to win. The player whose pioneer will be able to substitute all pairs of consecutive 1s by reinforced0s will come out winner.## Problem Overview:
## Implementation in C++:The mechanism function as an input, which is iterated by constant pairs of 0s till it converts to 1s. After that, we bring our made sequence to see who won.'determineWinner()'## Program:
Enter the sequence of 0s and 1s: 110101000110 Iteration 1: Found pair of consecutive 1s at index 0 and 1. Converting them to 0s. Found pair of consecutive 1s at index 9 and 10. Converting them to 0s. Updated sequence: 000101000000 Iteration 2: Updated sequence: 000101000000 Player 2 wins!
**Inclusion of Header Files:**
The code includes the execution direction (using iostream) and **Function determineWinner():**
The computer determines the of 1s can be placed in a consecutive sequence. Lastly, the winner is decided based on the element ofno pairs remaining blind spots.**Main Function:**
The main() function is responsible for directing the user to input the initial sequence of the binaries. The function **Detailed Comments:**
The comments within the code lean on every step of the procedure, such as initializing variables, looping through the sequence, and checking for **Output:**
In the middle of the function execution of , the position indices of the series of ones discovered in each iteration, and the sequence changed after the iteration.current versionThe code basically gives the player a more 1s with a single 0s. It comprises feedback and verbose data results that will help user to debug and get used to the grammar.pairs of consecutive## Complexity Analysis:
The time complexity of the code depends mostly on the length of the input sequence.
The algorithm repeats the tests on the where the input sequence consists only of 1s, the code will iterate more than once. It occurs with an asymptotic growth of unfavourable situation at each iteration of the sequence, where n is its length.O(n)
While being executed, the code iterates the sequence to find consecutive pairs of 1s within each iteration. This algorithm's , where n is the sequence length.O(n)Finally, the code complexity is that constantly loop through the sequence elements.nested loops
The code space complexity depends on the code's input sequence and the variables used to process them.
Input sequence string occupies
The code has a boolean variable called These variables occupy
On the other hand, this algorithm does not include recursive calls and no large function call insignificant.The code space complexity is and does not have a great influence on the overall space complexity.constant factor## Approach-2: Stack ApproachThe " . This method implements a stack data structure to maintain the indices of recorded '1's. During the sequence exploration, each '1' binary digits is pushed onto the stacking list. When repeated '1's are found, their indices are popped from the stack, and the characters in the sequence are replaced with '0's.encounteredSuch processing goes on until no more pairs of '1's adjacent to each other are produced. The Stack Strategy allows a more precise The time complexity is determined by the number of elements and doublets of '1's, while the size of the stack (in relation to the number of '1's) sets the space complexity.code modularity.## Program:
Enter the sequence of 0s and 1s: 1101001001 Final sequence: Converting pair of consecutive 1s at indices 0 and 1 to 0s. Converting pair of consecutive 1s at indices 3 and 6 to 0s. Player 2 wins! Total iterations: 2 0000000001
**Function determineWinner():**
This function takes as an that expresses the initial condition of game. It initializes a stack (std::stack<int> indexStack) for keeping track of the positions of the '1's. It repeats the sequence, moving the indices of the '1's to the stack.string sequenceIf two '1's are found in a row, it to '0's and increments the iteration counter. In an end, it checks if there are any '1's left to tell who the winner is: Player 1 if there are no '1's, otherwise Player 2. Eventually, it returns the transformed sequence.linked elements**Main Function:**
The user must input the initial sequence of 0s and 1s in the main () function. Finally, the determineWinner() function is called with the sequence as an input variable, and the It displays the informative console output, reflecting the process of substituting consecutive pairs of '1's to '0's and the winning party. The function ## Complexity Analysis:
The code loops as a single It is common for the duration of each loop to result
The loop carries out the operating task of removing and placing items into the stack using the Stack operations have in average a complexity of But, the number of
According to the stack-based mechanism, consecutive '1' pairs of the sequence must be retrieved. These methods of checks have a time complexity that is and accessing the elements using the comparisonsindex number.Considering these factors, the code time complexity will be ## Space Complexity:
The space complexity of where n is the length of the input sequence.O(n)
In the case of the code, the , and n is the input sequence.O(n) |

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