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Formatted question description: https://leetcode.ca/all/1475.html
1475. Final Prices With a Special Discount in a Shop (Easy)
Given the array prices
where prices[i]
is the price of the ith
item in a shop. There is a special discount for items in the shop, if you buy the ith
item, then you will receive a discount equivalent to prices[j]
where j
is the minimum index such that j > i
and prices[j] <= prices[i]
, otherwise, you will not receive any discount at all.
Return an array where the ith
element is the final price you will pay for the ith
item of the shop considering the special discount.
Example 1:
Input: prices = [8,4,6,2,3] Output: [4,2,4,2,3] Explanation: For item 0 with price[0]=8 you will receive a discount equivalent to prices[1]=4, therefore, the final price you will pay is 8  4 = 4. For item 1 with price[1]=4 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 4  2 = 2. For item 2 with price[2]=6 you will receive a discount equivalent to prices[3]=2, therefore, the final price you will pay is 6  2 = 4. For items 3 and 4 you will not receive any discount at all.
Example 2:
Input: prices = [1,2,3,4,5] Output: [1,2,3,4,5] Explanation: In this case, for all items, you will not receive any discount at all.
Example 3:
Input: prices = [10,1,1,6] Output: [9,0,1,6]
Constraints:
1 <= prices.length <= 500
1 <= prices[i] <= 10^3
Related Topics:
Array
Solution 1. Monotonic Stack
Similar idea as in 496. Next Greater Element I (Easy).
// OJ: https://leetcode.com/problems/finalpriceswithaspecialdiscountinashop/
// Time: O(N)
// Space: O(N)
class Solution {
public:
vector<int> finalPrices(vector<int>& A) {
stack<int> s;
int N = A.size();
for (int i = 0; i < N; ++i) {
while (s.size() && A[s.top()] >= A[i]) {
A[s.top()] = A[i];
s.pop();
}
s.push(i);
}
return A;
}
};

class Solution { public int[] finalPrices(int[] prices) { int length = prices.length; int[] finalPrices = new int[length]; for (int i = 0; i < length; i++) finalPrices[i] = prices[i]; for (int i = 0; i < length; i++) { for (int j = i + 1; j < length; j++) { if (prices[j] <= prices[i]) { finalPrices[i] = prices[j]; break; } } } return finalPrices; } }

// OJ: https://leetcode.com/problems/finalpriceswithaspecialdiscountinashop/ // Time: O(N) // Space: O(N) class Solution { public: vector<int> finalPrices(vector<int>& A) { stack<int> s; int N = A.size(); for (int i = 0; i < N; ++i) { while (s.size() && A[s.top()] >= A[i]) { A[s.top()] = A[i]; s.pop(); } s.push(i); } return A; } };

class Solution: def finalPrices(self, prices: List[int]) > List[int]: stk = [] ans = prices[:] for i, v in enumerate(prices): while stk and prices[stk[1]] >= v: ans[stk.pop()] = v stk.append(i) return ans