## 1475. Final Prices With a Special Discount in a Shop | LeetCode Solution

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`

class Solution

{

public:

vector<int> finalPrices(vector<int> &v)

{

stack<int> st;

int n = v.size();

vector<int> prev(n, -1);

for (int i = 0; i < n; i++)

{

while (!st.empty() && v[st.top()] >= v[i])

{

prev[st.top()] = i;

st.pop();

}

st.push(i);

}

for (int i = 0; i < n; i++)

{

if (prev[i] == -1)

prev[i] = v[i];

else

prev[i] = v[i] - v[prev[i]];

}

return prev;

}

};

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