LeetCode 3468: Find the Number of Copy Arrays Solution
Master LeetCode problem 3468 (Find the Number of Copy Arrays), 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 can iterate through the given original array and bounds array simultaneously. For each index i, we calculate the minimum and maximum possible values that can be copied at that index i based on the bounds. Then, we update these minimum and maximum values as we move along the array.
We can maintain a counter to keep track of the number of valid arrays that satisfy the condition mentioned in the problem.
Algorithm:
- Initialize a counter variable
countto 1 (initially considering the original array itself as a valid copy). - Iterate over the arrays using a loop from index 1 to n-1:
- Calculate the minimum and maximum values that can be copied at the current index based on the bounds.
- Update the counter
countby multiplying it with the number of valid values that can be copied at the current index.
- Return the final value of the counter
count.
Time Complexity:
The time complexity of this solution is O(n), where n is the length of the original array.
Space Complexity:
The space complexity of this solution is O(1) as we are using only a constant amount of extra space.
:
Solution Code
class Solution {
public int numOfArrays(int[] original, int[][] bounds) {
long count = 1;
int mod = 1000000007;
for (int i = 1; i < original.length; i++) {
int min = bounds[i][0];
int max = bounds[i][1];
count = (count * (Math.min(max, original[i]) - Math.max(min, original[i - 1]) + 1)) % mod;
if (count == 0) {
break;
}
}
return (int) count;
}
}Try It Yourself
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