Explanation

Given a list of customer orders, we need to find the number of customers who have strictly increasing purchases compared to the previous order. We can solve this problem by iterating through the customer orders and keeping track of the longest increasing subsequence ending at each order.

Here's the step-by-step algorithm:

Initialize an array dp of size equal to the number of customer orders, filled with 1s to represent the minimum subsequence length ending at each order.
Iterate through the customer orders starting from the second order.
For each order, compare it with all previous orders to find the longest increasing subsequence length ending at the current order.
Update the dp array with the maximum subsequence length found so far.
Return the sum of the dp array to get the total number of customers with strictly increasing purchases.

Time complexity: O(n^2) where n is the number of customer orders Space complexity: O(n) for the dp array

View Leetcode Problem

class Solution {
    public int maxProfit(int[] nums) {
        if (nums == null || nums.length == 0) {
            return 0;
        }
        
        int n = nums.length;
        int[] dp = new int[n];
        Arrays.fill(dp, 1);
        int maxProfit = 1;
        
        for (int i = 1; i < n; i++) {
            for (int j = 0; j < i; j++) {
                if (nums[i] > nums[j]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
            maxProfit = Math.max(maxProfit, dp[i]);
        }
        
        return maxProfit;
    }
}

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2474. Customers With Strictly Increasing Purchases

Explanation

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