LeetCode-in-Java
2931. Maximum Spending After Buying Items
Hard
You are given a 0-indexed m * n integer matrix values, representing the values of m * n different items in m different shops. Each shop has n items where the jth item in the ith shop has a value of values[i][j]. Additionally, the items in the ith shop are sorted in non-increasing order of value. That is, values[i][j] >= values[i][j + 1] for all 0 <= j < n - 1.
On each day, you would like to buy a single item from one of the shops. Specifically, On the dth day you can:
- Pick any shop
i. - Buy the rightmost available item
jfor the price ofvalues[i][j] * d. That is, find the greatest indexjsuch that itemjwas never bought before, and buy it for the price ofvalues[i][j] * d.
Note that all items are pairwise different. For example, if you have bought item 0 from shop 1, you can still buy item 0 from any other shop.
Return the maximum amount of money that can be spent on buying all m * n products.
Example 1:
Input: values = [[8,5,2],[6,4,1],[9,7,3]]
Output: 285
Explanation: On the first day, we buy product 2 from shop 1 for a price of values[1][2] * 1 = 1.
On the second day, we buy product 2 from shop 0 for a price of values[0][2] * 2 = 4.
On the third day, we buy product 2 from shop 2 for a price of values[2][2] * 3 = 9.
On the fourth day, we buy product 1 from shop 1 for a price of values[1][1] * 4 = 16.
On the fifth day, we buy product 1 from shop 0 for a price of values[0][1] * 5 = 25.
On the sixth day, we buy product 0 from shop 1 for a price of values[1][0] * 6 = 36.
On the seventh day, we buy product 1 from shop 2 for a price of values[2][1] * 7 = 49.
On the eighth day, we buy product 0 from shop 0 for a price of values[0][0] * 8 = 64.
On the ninth day, we buy product 0 from shop 2 for a price of values[2][0] * 9 = 81.
Hence, our total spending is equal to 285.
It can be shown that 285 is the maximum amount of money that can be spent buying all m * n products.
Example 2:
Input: values = [[10,8,6,4,2],[9,7,5,3,2]]
Output: 386
Explanation: On the first day, we buy product 4 from shop 0 for a price of values[0][4] * 1 = 2.
On the second day, we buy product 4 from shop 1 for a price of values[1][4] * 2 = 4.
On the third day, we buy product 3 from shop 1 for a price of values[1][3] * 3 = 9.
On the fourth day, we buy product 3 from shop 0 for a price of values[0][3] * 4 = 16.
On the fifth day, we buy product 2 from shop 1 for a price of values[1][2] * 5 = 25.
On the sixth day, we buy product 2 from shop 0 for a price of values[0][2] * 6 = 36.
On the seventh day, we buy product 1 from shop 1 for a price of values[1][1] * 7 = 49.
On the eighth day, we buy product 1 from shop 0 for a price of values[0][1] * 8 = 64
On the ninth day, we buy product 0 from shop 1 for a price of values[1][0] * 9 = 81.
On the tenth day, we buy product 0 from shop 0 for a price of values[0][0] * 10 = 100.
Hence, our total spending is equal to 386.
It can be shown that 386 is the maximum amount of money that can be spent buying all m * n products.
Constraints:
1 <= m == values.length <= 101 <= n == values[i].length <= 1041 <= values[i][j] <= 106values[i]are sorted in non-increasing order.
Solution
public class Solution {
private static class Node {
int val = -1;
Node next = null;
public Node(int val) {
this.val = val;
}
public Node() {}
}
public long maxSpending(int[][] values) {
int m = values.length;
int n = values[0].length;
Node head = new Node();
Node node = head;
for (int j = n - 1; j >= 0; j--) {
node.next = new Node(values[0][j]);
node = node.next;
}
for (int i = 1; i < m; i++) {
node = head;
for (int j = n - 1; j >= 0; j--) {
while (node.next != null && node.next.val <= values[i][j]) {
node = node.next;
}
Node next = node.next;
node.next = new Node(values[i][j]);
node = node.next;
node.next = next;
}
}
long res = 0;
long day = 1;
node = head.next;
while (node != null) {
res += day * node.val;
node = node.next;
day++;
}
return res;
}
}