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Sherlock and Cost
Sherlock and Cost
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testcase=int(input()) for i in range(testcase): n=int(input()) b=[int(x) for x in input().split(' ')]
maxi, max1 = 0, 0 for i in range(1,len(b)): curr, prev = b[i], b[i-1] newmaxi = max(maxi + abs(curr - prev), max1 + (curr - 1)) newmax1 = max(maxi + abs(1 - prev), max1) maxi, max1 = newmaxi, newmax1
print(max(maxi, max1))
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Here's a PHP solution to solve this problem:
Explanation:
Dynamic Programming Table (
dp
):dp[0]
: The maximum cost whenA[i]
is set to1
.dp[1]
: The maximum cost whenA[i]
is set toB[i]
.Transition between states:
i
from1
ton-1
, we updatedp[0]
anddp[1]
based on the previous values:A[i]
is1
, the cost can come from eitherA[i-1]
being1
orB[i-1]
.A[i]
isB[i]
, the cost can come from eitherA[i-1]
being1
orB[i-1]
.Final Result:
dp[0]
anddp[1]
at the end of the array, which represents the maximum cost achievable by either settingA[n-1]
to1
orB[n-1]
.This solution efficiently computes the desired result with a time complexity of
O(n)
, making it suitable for large inputs within the constraints.