Feat: Dynamic Programming

Cpp/Dynamic Programming/ClimbingStairs.cpp

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+#include<iostream>
+
+using namespace std;
+
+int ClimbStairs(int n);
+
+int main()
+{
+    cout << "Enter the number of stairs: ";
+    int n;
+    cin >> n;
+    cout << ClimbStairs(n) << endl;
+}
+
+int ClimbStairs(int n)
+{
+    int dp[n+1];
+    dp[0] = 1;     // If there the stairs are 0 and 1 in number, then the possible ways to climb are 1 and 1 respectively.
+    dp[1] = 1;
+    for(int i=2;i<=n;i++)
+    {
+        dp[i] = dp[i-1] + dp[i-2];     // To climb to nth stairs, the number of ways depends on the number of ways to climb (n-1)th and (n-2)th stairs
+    }
+    return dp[n];
+}
+
+  /* 
+
+    *** Test Case 1 ***
+
+    Enter the number of stairs: 6
+
+    13
+
+    *** Test Case 2 ***
+
+    Enter the number of stairs: 56
+
+    363076002
+
+    Link to the problem: https://leetcode.com/problems/climbing-stairs/
+
+   */

Cpp/Dynamic Programming/max_stock_profit.cpp

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+// Program to find the maximum profit when tradiing in a stock
+
+// PROBLEM STATEMENT
+
+/*
+    You are given an array prices where prices[i] is the price of a given stock on the ith day.
+
+    You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.   
+
+    Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.
+*/
+
+#include<iostream>
+#include<vector>
+
+using namespace std;
+
+int maxProfit(vector<int>&prices);
+
+// Driver Code
+
+int main()
+{
+    int n;
+    cin >> n;
+    vector<int> prices(n);
+    for(int i=0;i<n;i++)
+    {
+        cin >> prices[i];
+    }
+
+    cout << maxProfit(prices) << endl;;
+}
+
+// Implemented function for the solution.
+
+int maxProfit(vector<int> &prices)
+{
+    int n = prices.size();
+    int minPriceSoFar = INT32_MAX;  // Stores the minimum price of a stock till the selling day
+    int maxProfit = INT32_MIN;      // Stores the final answer of maximum profit.
+
+    for(int i=0;i<n;i++)
+    {
+        minPriceSoFar = min(minPriceSoFar,prices[i]);
+        int profit = prices[i] - minPriceSoFar;    //  Stores the profit for each day after selling.
+        maxProfit = max(maxProfit,profit);
+    }
+    return maxProfit;
+}