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47 changes: 47 additions & 0 deletions solutions/Firstsolution.txt
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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