LeetCode 1007: Minimum Domino Rotations For Equal Row Solution
Master LeetCode problem 1007 (Minimum Domino Rotations For Equal Row), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
Problem Explanation
Explanation
To solve this problem, we need to find the minimum number of rotations required to make all values in either the top row or the bottom row the same. We can achieve this by checking each possible value (from 1 to 6) and counting the number of rotations needed for each value.
- Iterate through the possible values from 1 to 6.
- For each value, count the number of rotations needed if we make all values in the top row equal to that value or all values in the bottom row equal to that value.
- Return the minimum number of rotations or -1 if it's not possible.
- Time complexity: O(n), where n is the length of the tops array.
- Space complexity: O(1).
Solution Code
class Solution {
public int minDominoRotations(int[] tops, int[] bottoms) {
int n = tops.length;
int[] topCounts = new int[7];
int[] bottomCounts = new int[7];
int[] sameCounts = new int[7];
for (int i = 0; i < n; i++) {
topCounts[tops[i]]++;
bottomCounts[bottoms[i]]++;
if (tops[i] == bottoms[i]) {
sameCounts[tops[i]]++;
}
}
for (int i = 1; i <= 6; i++) {
if (topCounts[i] + bottomCounts[i] - sameCounts[i] == n) {
return n - Math.max(topCounts[i], bottomCounts[i]);
}
}
return -1;
}
}Try It Yourself
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