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Formatted question description: https://leetcode.ca/all/1536.html
1536. Minimum Swaps to Arrange a Binary Grid (Medium)
Given an n x n
binary grid
, in one step you can choose two adjacent rows of the grid and swap them.
A grid is said to be valid if all the cells above the main diagonal are zeros.
Return the minimum number of steps needed to make the grid valid, or -1 if the grid cannot be valid.
The main diagonal of a grid is the diagonal that starts at cell (1, 1)
and ends at cell (n, n)
.
Example 1:
Input: grid = [[0,0,1],[1,1,0],[1,0,0]] Output: 3
Example 2:
Input: grid = [[0,1,1,0],[0,1,1,0],[0,1,1,0],[0,1,1,0]] Output: -1 Explanation: All rows are similar, swaps have no effect on the grid.
Example 3:
Input: grid = [[1,0,0],[1,1,0],[1,1,1]] Output: 0
Constraints:
n == grid.length
n == grid[i].length
1 <= n <= 200
grid[i][j]
is0
or1
Related Topics:
Greedy
Solution 1.
First store the max index of the 1
in each row into a vector<int> v
.
Then we can from the first item to the last item in v
:
- if
v[i] <= i
, this row is good. Skip - Otherwise, we find the minimal
j
(i + 1 < j < N
) that satisfiesv[j] <= i
.- If we can’t find it, no row can be placed at this
i
-th row. Return-1
. - Otherwise, we move
v[j]
tov[i]
and thev[k]
(i <= k < j
) are moved downwards. And we addj - i
steps of swaps to the answer.
- If we can’t find it, no row can be placed at this
// https://leetcode.com/problems/minimum-swaps-to-arrange-a-binary-grid/
// Time: O(N^2)
// Space: O(N)
class Solution {
public:
int minSwaps(vector<vector<int>>& G) {
int N = G.size(), ans = 0;
vector<int> v(N);
for (int i = 0; i < N; ++i) {
int len = 0;
for (int j = N - 1; j >= 0; --j) {
if (G[i][j] == 0) continue;
len = j + 1;
break;
}
v[i] = len;
}
for (int i = 0; i < N; ++i) {
if (v[i] <= i + 1) continue;
int j = i + 1;
while (j < N && v[j] > i + 1) ++j;
if (j == N) return -1;
int tmp = v[j];
for (int k = j; k > i; --k) v[k] = v[k - 1];
v[i] = tmp;
ans += j - i;
}
return ans;
}
};
-
class Solution { public int minSwaps(int[][] grid) { int side = grid.length; int[] zerosCounts = new int[side]; for (int i = 0; i < side; i++) { int[] row = grid[i]; int zerosCount = 0; for (int j = side - 1; j >= 0; j--) { if (row[j] == 0) zerosCount++; else break; } zerosCounts[i] = zerosCount; } int swaps = 0; for (int i = 0; i < side; i++) { int minZeros = side - i - 1; if (zerosCounts[i] < minZeros) { int firstRowIndex = -1; for (int j = i + 1; j < side; j++) { if (zerosCounts[j] >= minZeros) { firstRowIndex = j; break; } } if (firstRowIndex < 0) return -1; for (int j = firstRowIndex; j > i; j--) { swaps++; int temp = zerosCounts[j]; zerosCounts[j] = zerosCounts[j - 1]; zerosCounts[j - 1] = temp; } } } return swaps; } }
-
// https://leetcode.com/problems/minimum-swaps-to-arrange-a-binary-grid/ // Time: O(N^2) // Space: O(N) class Solution { public: int minSwaps(vector<vector<int>>& G) { int N = G.size(), ans = 0; vector<int> v(N); for (int i = 0; i < N; ++i) { int len = 0; for (int j = N - 1; j >= 0; --j) { if (G[i][j] == 0) continue; len = j + 1; break; } v[i] = len; } for (int i = 0; i < N; ++i) { if (v[i] <= i + 1) continue; int j = i + 1; while (j < N && v[j] > i + 1) ++j; if (j == N) return -1; int tmp = v[j]; for (int k = j; k > i; --k) v[k] = v[k - 1]; v[i] = tmp; ans += j - i; } return ans; } };