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2297. Jump Game VIII
Description
You are given a 0-indexed integer array nums
of length n
. You are initially standing at index 0
. You can jump from index i
to index j
where i < j
if:
nums[i] <= nums[j]
andnums[k] < nums[i]
for all indexesk
in the rangei < k < j
, ornums[i] > nums[j]
andnums[k] >= nums[i]
for all indexesk
in the rangei < k < j
.
You are also given an integer array costs
of length n
where costs[i]
denotes the cost of jumping to index i
.
Return the minimum cost to jump to the index n - 1
.
Example 1:
Input: nums = [3,2,4,4,1], costs = [3,7,6,4,2] Output: 8 Explanation: You start at index 0. - Jump to index 2 with a cost of costs[2] = 6. - Jump to index 4 with a cost of costs[4] = 2. The total cost is 8. It can be proven that 8 is the minimum cost needed. Two other possible paths are from index 0 -> 1 -> 4 and index 0 -> 2 -> 3 -> 4. These have a total cost of 9 and 12, respectively.
Example 2:
Input: nums = [0,1,2], costs = [1,1,1] Output: 2 Explanation: Start at index 0. - Jump to index 1 with a cost of costs[1] = 1. - Jump to index 2 with a cost of costs[2] = 1. The total cost is 2. Note that you cannot jump directly from index 0 to index 2 because nums[0] <= nums[1].
Constraints:
n == nums.length == costs.length
1 <= n <= 105
0 <= nums[i], costs[i] <= 105
Solutions
-
class Solution { public long minCost(int[] nums, int[] costs) { int n = nums.length; List<Integer>[] g = new List[n]; Arrays.setAll(g, k -> new ArrayList<>()); Deque<Integer> stk = new ArrayDeque<>(); for (int i = n - 1; i >= 0; --i) { while (!stk.isEmpty() && nums[stk.peek()] < nums[i]) { stk.pop(); } if (!stk.isEmpty()) { g[i].add(stk.peek()); } stk.push(i); } stk.clear(); for (int i = n - 1; i >= 0; --i) { while (!stk.isEmpty() && nums[stk.peek()] >= nums[i]) { stk.pop(); } if (!stk.isEmpty()) { g[i].add(stk.peek()); } stk.push(i); } long[] f = new long[n]; Arrays.fill(f, 1L << 60); f[0] = 0; for (int i = 0; i < n; ++i) { for (int j : g[i]) { f[j] = Math.min(f[j], f[i] + costs[j]); } } return f[n - 1]; } }
-
class Solution { public: long long minCost(vector<int>& nums, vector<int>& costs) { int n = nums.size(); vector<int> g[n]; stack<int> stk; for (int i = n - 1; ~i; --i) { while (!stk.empty() && nums[stk.top()] < nums[i]) { stk.pop(); } if (!stk.empty()) { g[i].push_back(stk.top()); } stk.push(i); } stk = stack<int>(); for (int i = n - 1; ~i; --i) { while (!stk.empty() && nums[stk.top()] >= nums[i]) { stk.pop(); } if (!stk.empty()) { g[i].push_back(stk.top()); } stk.push(i); } vector<long long> f(n, 1e18); f[0] = 0; for (int i = 0; i < n; ++i) { for (int j : g[i]) { f[j] = min(f[j], f[i] + costs[j]); } } return f[n - 1]; } };
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class Solution: def minCost(self, nums: List[int], costs: List[int]) -> int: n = len(nums) g = defaultdict(list) stk = [] for i in range(n - 1, -1, -1): while stk and nums[stk[-1]] < nums[i]: stk.pop() if stk: g[i].append(stk[-1]) stk.append(i) stk = [] for i in range(n - 1, -1, -1): while stk and nums[stk[-1]] >= nums[i]: stk.pop() if stk: g[i].append(stk[-1]) stk.append(i) f = [inf] * n f[0] = 0 for i in range(n): for j in g[i]: f[j] = min(f[j], f[i] + costs[j]) return f[n - 1]
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func minCost(nums []int, costs []int) int64 { n := len(nums) g := make([][]int, n) stk := []int{} for i := n - 1; i >= 0; i-- { for len(stk) > 0 && nums[stk[len(stk)-1]] < nums[i] { stk = stk[:len(stk)-1] } if len(stk) > 0 { g[i] = append(g[i], stk[len(stk)-1]) } stk = append(stk, i) } stk = []int{} for i := n - 1; i >= 0; i-- { for len(stk) > 0 && nums[stk[len(stk)-1]] >= nums[i] { stk = stk[:len(stk)-1] } if len(stk) > 0 { g[i] = append(g[i], stk[len(stk)-1]) } stk = append(stk, i) } f := make([]int64, n) for i := 1; i < n; i++ { f[i] = math.MaxInt64 } for i := 0; i < n; i++ { for _, j := range g[i] { f[j] = min(f[j], f[i]+int64(costs[j])) } } return f[n-1] }
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function minCost(nums: number[], costs: number[]): number { const n = nums.length; const g: number[][] = Array.from({ length: n }, () => []); const stk: number[] = []; for (let i = n - 1; i >= 0; --i) { while (stk.length && nums[stk[stk.length - 1]] < nums[i]) { stk.pop(); } if (stk.length) { g[i].push(stk[stk.length - 1]); } stk.push(i); } stk.length = 0; for (let i = n - 1; i >= 0; --i) { while (stk.length && nums[stk[stk.length - 1]] >= nums[i]) { stk.pop(); } if (stk.length) { g[i].push(stk[stk.length - 1]); } stk.push(i); } const f: number[] = Array.from({ length: n }, () => Infinity); f[0] = 0; for (let i = 0; i < n; ++i) { for (const j of g[i]) { f[j] = Math.min(f[j], f[i] + costs[j]); } } return f[n - 1]; }