From 2c5171f20286e198fa4b292b01be61db66da587d Mon Sep 17 00:00:00 2001 From: KHALID BOUASSA <69470548+khalidbouassa@users.noreply.github.com> Date: Mon, 26 Dec 2022 17:33:52 +0100 Subject: [PATCH 1/3] Create khalidBouassaSolution.txt --- khalidBouassaSolution.txt | 48 +++++++++++++++++++++++++++++++++++++++ 1 file changed, 48 insertions(+) create mode 100644 khalidBouassaSolution.txt diff --git a/khalidBouassaSolution.txt b/khalidBouassaSolution.txt new file mode 100644 index 0000000..85c80b7 --- /dev/null +++ b/khalidBouassaSolution.txt @@ -0,0 +1,48 @@ +The "Thirsty Men Problem" involves a group of men who are stranded in the desert and must decide how to ration a finite supply of water. +The problem can be stated as follows: + +There are 'n' men and 'm' jugs of water. Each man has a certain thirst level, represented by a positive integer. +A man can drink from a jug to decrease his thirst by an amount equal to the amount of water in the jug. However, once a jug is empty, it cannot be refilled. +The goal is to determine the minimum amount of water that must be consumed in order to satisfy the thirst of all n men. + +One approach to solving this problem is to use a greedy algorithm. The idea is to start by having each man drink from the jug with the most water first. +This will satisfy the thirst of the most men possible, and will also minimize the amount of water wasted. + +To implement this algorithm, we can use the following steps: + +Sort the jugs of water in descending order by the amount of water they contain. +For each man, starting with the thirstiest: +Have the man drink from the jug with the most water first. +If the man's thirst is not fully satisfied, move on to the next-largest jug and repeat until the man's thirst is satisfied. +Repeat this process for all n men. +This algorithm is guaranteed to find a solution that satisfies the thirst of all n men, as long as the total amount of water in the jugs is sufficient. It is also efficient, with a time complexity of O(n * m log m) (assuming a stable sort algorithm is used to sort the jugs). + +Here is some example code in Python that demonstrates how this algorithm could be implemented: + +def thirsty_men(n: int, m: int, thirst: List[int], jugs: List[int]) -> int: + # Sort the jugs in descending order by the amount of water they contain + jugs.sort(reverse=True) + + # Keep track of the total amount of water consumed + total_water = 0 + + # For each man, starting with the thirstiest: + for i in range(n): + # Have the man drink from the jug with the most water first + for j in range(m): + if thirst[i] == 0: + # The man's thirst is satisfied, move on to the next man + break + elif thirst[i] > jugs[j]: + # The man's thirst is not fully satisfied, but the jug is empty + thirst[i] -= jugs[j] + total_water += jugs[j] + jugs[j] = 0 + else: + # The man's thirst is fully satisfied + total_water += thirst[i] + jugs[j] -= thirst[i] + thirst[i] = 0 + break + + return total_water From e151b0683138673750f452a84367e7a9ef089d4e Mon Sep 17 00:00:00 2001 From: KHALID BOUASSA <69470548+khalidbouassa@users.noreply.github.com> Date: Mon, 26 Dec 2022 17:36:07 +0100 Subject: [PATCH 2/3] Update Firstsolution.txt --- solutions/Firstsolution.txt | 47 +++++++++++++++++++++++++++++++++++++ 1 file changed, 47 insertions(+) diff --git a/solutions/Firstsolution.txt b/solutions/Firstsolution.txt index 8b13789..85c80b7 100644 --- a/solutions/Firstsolution.txt +++ b/solutions/Firstsolution.txt @@ -1 +1,48 @@ +The "Thirsty Men Problem" involves a group of men who are stranded in the desert and must decide how to ration a finite supply of water. +The problem can be stated as follows: +There are 'n' men and 'm' jugs of water. Each man has a certain thirst level, represented by a positive integer. +A man can drink from a jug to decrease his thirst by an amount equal to the amount of water in the jug. However, once a jug is empty, it cannot be refilled. +The goal is to determine the minimum amount of water that must be consumed in order to satisfy the thirst of all n men. + +One approach to solving this problem is to use a greedy algorithm. The idea is to start by having each man drink from the jug with the most water first. +This will satisfy the thirst of the most men possible, and will also minimize the amount of water wasted. + +To implement this algorithm, we can use the following steps: + +Sort the jugs of water in descending order by the amount of water they contain. +For each man, starting with the thirstiest: +Have the man drink from the jug with the most water first. +If the man's thirst is not fully satisfied, move on to the next-largest jug and repeat until the man's thirst is satisfied. +Repeat this process for all n men. +This algorithm is guaranteed to find a solution that satisfies the thirst of all n men, as long as the total amount of water in the jugs is sufficient. It is also efficient, with a time complexity of O(n * m log m) (assuming a stable sort algorithm is used to sort the jugs). + +Here is some example code in Python that demonstrates how this algorithm could be implemented: + +def thirsty_men(n: int, m: int, thirst: List[int], jugs: List[int]) -> int: + # Sort the jugs in descending order by the amount of water they contain + jugs.sort(reverse=True) + + # Keep track of the total amount of water consumed + total_water = 0 + + # For each man, starting with the thirstiest: + for i in range(n): + # Have the man drink from the jug with the most water first + for j in range(m): + if thirst[i] == 0: + # The man's thirst is satisfied, move on to the next man + break + elif thirst[i] > jugs[j]: + # The man's thirst is not fully satisfied, but the jug is empty + thirst[i] -= jugs[j] + total_water += jugs[j] + jugs[j] = 0 + else: + # The man's thirst is fully satisfied + total_water += thirst[i] + jugs[j] -= thirst[i] + thirst[i] = 0 + break + + return total_water From 23bd92d27fa74e77b7a8d504093b86412f40e050 Mon Sep 17 00:00:00 2001 From: KHALID BOUASSA <69470548+khalidbouassa@users.noreply.github.com> Date: Mon, 26 Dec 2022 17:36:22 +0100 Subject: [PATCH 3/3] Delete khalidBouassaSolution.txt --- khalidBouassaSolution.txt | 48 --------------------------------------- 1 file changed, 48 deletions(-) delete mode 100644 khalidBouassaSolution.txt diff --git a/khalidBouassaSolution.txt b/khalidBouassaSolution.txt deleted file mode 100644 index 85c80b7..0000000 --- a/khalidBouassaSolution.txt +++ /dev/null @@ -1,48 +0,0 @@ -The "Thirsty Men Problem" involves a group of men who are stranded in the desert and must decide how to ration a finite supply of water. -The problem can be stated as follows: - -There are 'n' men and 'm' jugs of water. Each man has a certain thirst level, represented by a positive integer. -A man can drink from a jug to decrease his thirst by an amount equal to the amount of water in the jug. However, once a jug is empty, it cannot be refilled. -The goal is to determine the minimum amount of water that must be consumed in order to satisfy the thirst of all n men. - -One approach to solving this problem is to use a greedy algorithm. The idea is to start by having each man drink from the jug with the most water first. -This will satisfy the thirst of the most men possible, and will also minimize the amount of water wasted. - -To implement this algorithm, we can use the following steps: - -Sort the jugs of water in descending order by the amount of water they contain. -For each man, starting with the thirstiest: -Have the man drink from the jug with the most water first. -If the man's thirst is not fully satisfied, move on to the next-largest jug and repeat until the man's thirst is satisfied. -Repeat this process for all n men. -This algorithm is guaranteed to find a solution that satisfies the thirst of all n men, as long as the total amount of water in the jugs is sufficient. It is also efficient, with a time complexity of O(n * m log m) (assuming a stable sort algorithm is used to sort the jugs). - -Here is some example code in Python that demonstrates how this algorithm could be implemented: - -def thirsty_men(n: int, m: int, thirst: List[int], jugs: List[int]) -> int: - # Sort the jugs in descending order by the amount of water they contain - jugs.sort(reverse=True) - - # Keep track of the total amount of water consumed - total_water = 0 - - # For each man, starting with the thirstiest: - for i in range(n): - # Have the man drink from the jug with the most water first - for j in range(m): - if thirst[i] == 0: - # The man's thirst is satisfied, move on to the next man - break - elif thirst[i] > jugs[j]: - # The man's thirst is not fully satisfied, but the jug is empty - thirst[i] -= jugs[j] - total_water += jugs[j] - jugs[j] = 0 - else: - # The man's thirst is fully satisfied - total_water += thirst[i] - jugs[j] -= thirst[i] - thirst[i] = 0 - break - - return total_water