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

## 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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