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1473. Paint House III

There is a row of m houses in a small city, each house must be painted with one of the n colors (labeled from 1 to n), some houses that have been painted last summer should not be painted again.

A neighborhood is a maximal group of continuous houses that are painted with the same color.

  • For example: houses = [1,2,2,3,3,2,1,1] contains 5 neighborhoods [{1}, {2,2}, {3,3}, {2}, {1,1}].

Given an array houses, an m x n matrix cost and an integer target where:

  • houses[i]: is the color of the house i, and 0 if the house is not painted yet.
  • cost[i][j]: is the cost of paint the house i with the color j + 1.

Return the minimum cost of painting all the remaining houses in such a way that there are exactlytargetneighborhoods. If it is not possible, return -1.

Example 1:

Input: houses = [0,0,0,0,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3 Output: 9 Explanation: Paint houses of this way [1,2,2,1,1] This array contains target = 3 neighborhoods, [{1}, {2,2}, {1,1}]. Cost of paint all houses (1 + 1 + 1 + 1 + 5) = 9.

Example 2:

Input: houses = [0,2,1,2,0], cost = [[1,10],[10,1],[10,1],[1,10],[5,1]], m = 5, n = 2, target = 3 Output: 11 Explanation: Some houses are already painted, Paint the houses of this way [2,2,1,2,2] This array contains target = 3 neighborhoods, [{2,2}, {1}, {2,2}]. Cost of paint the first and last house (10 + 1) = 11.

Example 3:

Input: houses = [3,1,2,3], cost = [[1,1,1],[1,1,1],[1,1,1],[1,1,1]], m = 4, n = 3, target = 3 Output: -1 Explanation: Houses are already painted with a total of 4 neighborhoods [{3},{1},{2},{3}] different of target = 3.

Constraints:

  • m == houses.length == cost.length
  • n == cost[i].length
  • 1 <= m <= 100
  • 1 <= n <= 20
  • 1 <= target <= m
  • 0 <= houses[i] <= n
  • 1 <= cost[i][j] <= 104

Solutions (Rust)

1. Solution

implSolution{pubfnmin_cost(houses:Vec<i32>,cost:Vec<Vec<i32>>,m:i32,n:i32,target:i32) -> i32{let(m, n, target) = (m asusize, n asusize, target asusize);let houses = houses.into_iter().map(|x| x asusize).collect::<Vec<_>>();letmut dp = vec![vec![vec![None; n]; target + 1]; m];if houses[0] > 0{ dp[0][1][houses[0]asusize - 1] = Some(0);}else{for j in0..n { dp[0][1][j] = Some(cost[0][j]);}}for i in1..m {for k in1..=target {for j in0..n {ifletSome(x) = dp[i - 1][k][j]{if houses[i] > 0{if houses[i] - 1 == j { dp[i][k][j] = Some(dp[i][k][j].unwrap_or(i32::MAX).min(x));}elseif k + 1 <= target { dp[i][k + 1][houses[i] - 1] = Some(dp[i][k + 1][houses[i] - 1].unwrap_or(i32::MAX).min(x));}}else{for jj in0..n {if jj == j { dp[i][k][jj] = Some(dp[i][k][jj].unwrap_or(i32::MAX).min(x + cost[i][jj]));}elseif k + 1 <= target { dp[i][k + 1][jj] = Some( dp[i][k + 1][jj].unwrap_or(i32::MAX).min(x + cost[i][jj]),);}}}}}}} dp[m - 1][target].iter().filter_map(|&x| x).min().unwrap_or(-1)}}
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