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2162. Minimum Cost to Set Cooking Time
Description
A generic microwave supports cooking times for:
 at least
1
second.  at most
99
minutes and99
seconds.
To set the cooking time, you push at most four digits. The microwave normalizes what you push as four digits by prepending zeroes. It interprets the first two digits as the minutes and the last two digits as the seconds. It then adds them up as the cooking time. For example,
 You push
9
5
4
(three digits). It is normalized as0954
and interpreted as9
minutes and54
seconds.  You push
0
0
0
8
(four digits). It is interpreted as0
minutes and8
seconds.  You push
8
0
9
0
. It is interpreted as80
minutes and90
seconds.  You push
8
1
3
0
. It is interpreted as81
minutes and30
seconds.
You are given integers startAt
, moveCost
, pushCost
, and targetSeconds
. Initially, your finger is on the digit startAt
. Moving the finger above any specific digit costs moveCost
units of fatigue. Pushing the digit below the finger once costs pushCost
units of fatigue.
There can be multiple ways to set the microwave to cook for targetSeconds
seconds but you are interested in the way with the minimum cost.
Return the minimum cost to set targetSeconds
seconds of cooking time.
Remember that one minute consists of 60
seconds.
Example 1:
Input: startAt = 1, moveCost = 2, pushCost = 1, targetSeconds = 600 Output: 6 Explanation: The following are the possible ways to set the cooking time.  1 0 0 0, interpreted as 10 minutes and 0 seconds. The finger is already on digit 1, pushes 1 (with cost 1), moves to 0 (with cost 2), pushes 0 (with cost 1), pushes 0 (with cost 1), and pushes 0 (with cost 1). The cost is: 1 + 2 + 1 + 1 + 1 = 6. This is the minimum cost.  0 9 6 0, interpreted as 9 minutes and 60 seconds. That is also 600 seconds. The finger moves to 0 (with cost 2), pushes 0 (with cost 1), moves to 9 (with cost 2), pushes 9 (with cost 1), moves to 6 (with cost 2), pushes 6 (with cost 1), moves to 0 (with cost 2), and pushes 0 (with cost 1). The cost is: 2 + 1 + 2 + 1 + 2 + 1 + 2 + 1 = 12.  9 6 0, normalized as 0960 and interpreted as 9 minutes and 60 seconds. The finger moves to 9 (with cost 2), pushes 9 (with cost 1), moves to 6 (with cost 2), pushes 6 (with cost 1), moves to 0 (with cost 2), and pushes 0 (with cost 1). The cost is: 2 + 1 + 2 + 1 + 2 + 1 = 9.
Example 2:
Input: startAt = 0, moveCost = 1, pushCost = 2, targetSeconds = 76 Output: 6 Explanation: The optimal way is to push two digits: 7 6, interpreted as 76 seconds. The finger moves to 7 (with cost 1), pushes 7 (with cost 2), moves to 6 (with cost 1), and pushes 6 (with cost 2). The total cost is: 1 + 2 + 1 + 2 = 6 Note other possible ways are 0076, 076, 0116, and 116, but none of them produces the minimum cost.
Constraints:
0 <= startAt <= 9
1 <= moveCost, pushCost <= 10^{5}
1 <= targetSeconds <= 6039
Solutions

class Solution { public int minCostSetTime(int startAt, int moveCost, int pushCost, int targetSeconds) { int m = targetSeconds / 60; int s = targetSeconds % 60; return Math.min( f(m, s, startAt, moveCost, pushCost), f(m  1, s + 60, startAt, moveCost, pushCost)); } private int f(int m, int s, int prev, int moveCost, int pushCost) { if (m < 0  m > 99  s < 0  s > 99) { return Integer.MAX_VALUE; } int[] arr = new int[] {m / 10, m % 10, s / 10, s % 10}; int i = 0; for (; i < 4 && arr[i] == 0; ++i) ; int t = 0; for (; i < 4; ++i) { if (arr[i] != prev) { t += moveCost; } t += pushCost; prev = arr[i]; } return t; } }

class Solution { public: int minCostSetTime(int startAt, int moveCost, int pushCost, int targetSeconds) { int m = targetSeconds / 60, s = targetSeconds % 60; return min(f(m, s, startAt, moveCost, pushCost), f(m  1, s + 60, startAt, moveCost, pushCost)); } int f(int m, int s, int prev, int moveCost, int pushCost) { if (m < 0  m > 99  s < 0  s > 99) return INT_MAX; vector<int> arr = {m / 10, m % 10, s / 10, s % 10}; int i = 0; for (; i < 4 && arr[i] == 0; ++i) ; int t = 0; for (; i < 4; ++i) { if (arr[i] != prev) t += moveCost; t += pushCost; prev = arr[i]; } return t; } };

class Solution: def minCostSetTime( self, startAt: int, moveCost: int, pushCost: int, targetSeconds: int ) > int: def f(m, s): if not 0 <= m < 100 or not 0 <= s < 100: return inf arr = [m // 10, m % 10, s // 10, s % 10] i = 0 while i < 4 and arr[i] == 0: i += 1 t = 0 prev = startAt for v in arr[i:]: if v != prev: t += moveCost t += pushCost prev = v return t m, s = divmod(targetSeconds, 60) ans = min(f(m, s), f(m  1, s + 60)) return ans

func minCostSetTime(startAt int, moveCost int, pushCost int, targetSeconds int) int { m, s := targetSeconds/60, targetSeconds%60 f := func(m, s int) int { if m < 0  m > 99  s < 0  s > 99 { return 0x3f3f3f3f } arr := []int{m / 10, m % 10, s / 10, s % 10} i := 0 for ; i < 4 && arr[i] == 0; i++ { } t := 0 prev := startAt for ; i < 4; i++ { if arr[i] != prev { t += moveCost } t += pushCost prev = arr[i] } return t } return min(f(m, s), f(m1, s+60)) }