Algorithm
Problem Name: 1007. Minimum Domino Rotations For Equal Row
In a row of dominoes, tops[i] and bottoms[i] represent the top and bottom halves of the ith domino. (A domino is a tile with two numbers from 1 to 6 - one on each half of the tile.)
We may rotate the ith domino, so that tops[i] and bottoms[i] swap values.
Return the minimum number of rotations so that all the values in tops are the same, or all the values in bottoms are the same.
If it cannot be done, return -1.
Example 1:
Input: tops = [2,1,2,4,2,2], bottoms = [5,2,6,2,3,2] Output: 2 Explanation: The first figure represents the dominoes as given by tops and bottoms: before we do any rotations. If we rotate the second and fourth dominoes, we can make every value in the top row equal to 2, as indicated by the second figure.
Example 2:
Input: tops = [3,5,1,2,3], bottoms = [3,6,3,3,4] Output: -1 Explanation: In this case, it is not possible to rotate the dominoes to make one row of values equal.
Constraints:
2 <= tops.length <= 2 * 104bottoms.length == tops.length1 <= tops[i], bottoms[i] <= 6
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int minDominoRotations(int[] A, int[] B) {
int[] counterA = new int[7];
int[] counterB = new int[7];
int[] same = new int[7];
int n = A.length;
for (int i = 0; i < n; i++) {
counterA[A[i]]++;
counterB[B[i]]++;
if (A[i] == B[i]) {
same[A[i]]++;
}
}
for (int i = 1; i < 7; i++) {
if (counterA[i] + counterB[i] - same[i] == n) {
return n - Math.max(counterA[i], counterB[i]);
}
}
return -1;
}
}
Input
Output
#2 Code Example with Python Programming
Code -
Python Programming
class Solution:
def minDominoRotations(self, A: List[int], B: List[int]) -> int:
res = min(len(A) - max(A.count(c), B.count(c)) if all(a == c or b == c for a, b in zip(A, B)) else float('inf') for c in (A[0], B[0]))
return res if res < float('inf') else -1
Input
Output
#3 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _1007_MinimumDominoRotationsForEqualRow
{
public int MinDominoRotations(int[] A, int[] B)
{
var result = CheckMinDominoRotations(A[0], A, B);
if (result != -1 || A[0] == B[0]) return result;
return CheckMinDominoRotations(B[0], A, B);
}
public int CheckMinDominoRotations(int x, int[] A, int[] B)
{
int rotateA = 0, rotateB = 0;
for (int i = 0; i < A.Length; i++)
{
if (A[i] != x && B[i] != x) return -1;
else if (A[i] != x) rotateA++;
else if (B[i] != x) rotateB++;
}
return Math.Min(rotateA, rotateB);
}
}
}
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Output
#4 Code Example with C Programming
Code -
C Programming
int minDominoRotations(int* A, int ASize, int* B, int BSize){
bool a_ok, b_ok;
int i, a_up, a_down, b_down, b_up;
a_ok = b_ok = true;
a_up = a_down = b_down = b_up = 0;
for (i = 0; i < ASize; i ++) {
if (a_ok && A[i] != A[0] && B[i] != A[0]) a_ok = false;
if (b_ok && B[i] != B[0] && A[i] != B[0]) b_ok = false;
if (!a_ok && !b_ok) return -1;
if (a_ok) {
if (A[i] != A[0]) a_up ++; // flip up to make all are A[0] on top
else if (B[i] != A[0]) a_down ++; // flip down to make all are A[0] in bottom
}
if (b_ok) {
if (B[i] != B[0]) b_down ++;
else if (A[i] != B[0]) b_up ++;
}
}
if (a_ok) return a_up < a_down ? a_up : a_down;
if (b_ok) return b_up < b_down ? b_up : b_down;
return -1;
}
Input
Output