2280. Minimum Lines to Represent a Line Chart
Explanation:
To solve this problem, we can iterate through the given array of stock prices and find the points where the price changes. Whenever the price changes, we increment the count of lines needed to represent the line chart. The minimum number of lines needed will be the count of these points where the price changes.
- Initialize a variable
linesto store the count of lines needed. - Iterate through the stock prices array and compare the price of the current day with the price of the previous day.
- If the price changes, increment the
linesby 1. - Return the final value of
linesas the minimum number of lines needed.
Time Complexity:
The time complexity of this algorithm is O(n) where n is the number of stock prices.
Space Complexity:
The space complexity of this algorithm is O(1) as we are using only a constant amount of extra space.
:
class Solution {
public int minLines(int[][] stockPrices) {
if (stockPrices.length == 0) {
return 0;
}
int lines = 1;
for (int i = 1; i < stockPrices.length; i++) {
if (stockPrices[i][1] != stockPrices[i - 1][1]) {
lines++;
}
}
return lines;
}
}Code Editor (Testing phase)
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