There are a row of n houses, each house can be painted with one of the k colors. The cost of painting each house with a certain color is different. You have to paint all the houses such that no two adjacent houses have the same color.
The cost of painting each house with a certain color is represented by an n x k cost matrix costs.
For example, costs[0][0] is the cost of painting house 0 with color 0; costs[1][2] is the cost of painting house 1 with color 2, and so on...
Return the minimum cost to paint all houses.
Example 1:
**Input:** costs = [[1,5,3],[2,9,4]]
**Output:** 5
**Explanation:**
Paint house 0 into color 0, paint house 1 into color 2. Minimum cost: 1 + 4 = 5;
Or paint house 0 into color 2, paint house 1 into color 0. Minimum cost: 3 + 2 = 5.
Example 2:
**Input:** costs = [[1,3],[2,4]]
**Output:** 5
Constraints:
costs.length == n
costs[i].length == k
1 <= n <= 100
2 <= k <= 20
1 <= costs[i][j] <= 20
Follow up: Could you solve it in O(nk) runtime?
ac
O(nk^2)
class Solution {
public int minCostII(int[][] costs) {
// edge cases
if (costs == null || costs.length == 0 || costs[0].length == 0)
return 0;
int row = costs.length;
int col = costs[0].length;
for (int r = 1; r < row; r++) {
for (int c = 0; c < col; c++) {
int min = Integer.MAX_VALUE; // min value from previous house
for (int k = 0; k < col; k++) {
if (k == c) continue; // avoid same color
min = Math.min(min, costs[r-1][k]);
}
costs[r][c] += min; // update curr cell
}
}
// check last row, get result
int res = Integer.MAX_VALUE;
for (int c = 0; c < col; c++) {
res = Math.min(res, costs[row-1][c]);
}
return res;
}
}
/*
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*/
class Solution {
public int minCostII(int[][] costs) {
// edge cases
if (costs == null || costs.length == 0 || costs[0].length == 0)
return 0;
int row = costs.length;
int col = costs[0].length;
int min1idx = -1, min2idx = -1;
for (int r = 0; r < row; r++) {
int tmpmin1 = -1, tmpmin2 = -1;
for (int c = 0; c < col; c++) {
if (r > 0) { // skip first row
if (min1idx == c) { // 1st min is same color, use 2nd color
costs[r][c] += costs[r-1][min2idx];
} else {
costs[r][c] += costs[r-1][min1idx];
}
}
// update 1st min and 2nd min
if (tmpmin1 == -1 || costs[r][c] < costs[r][tmpmin1]) {
tmpmin2 = tmpmin1;
tmpmin1 = c;
} else if (tmpmin2 == -1 || costs[r][c] < costs[r][tmpmin2]) {
tmpmin2 = c;
}
}
min1idx = tmpmin1;
min2idx = tmpmin2;
}
// check last row, get result
int res = Integer.MAX_VALUE;
for (int c = 0; c < col; c++) {
res = Math.min(res, costs[row-1][c]);
}
return res;
}
}
/*
from O(nk^2) to O(nk), keep 2 variables from previous step: 1st min, 2nd min
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*/