Day-15 stock span problem

Author: mandliya

stack_problems/README.md

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+##Problems solved using stack
+1. A simple demo of stack
+2. Stock Span Problem
+    Given : We have series of n daily price quotes for a stock. 
+    Required: We need to calculate **span** of stock's price for all n days.
+              Span for ith day is defined as **maximum** number of consecutive days,
+              for which the price of the stock was less than or equal to ith day.
+    Example: For stock quotes {100, 60, 70, 65, 80, 85} span will be {1, 1, 2, 1, 4, 5}.
+             Span for day 1 is always 1, now for day 2 stock is at 60, and there is no day befor it when stock was less than 60.
+             So span remains 1. For day 3, the stock is priced at 70, so its span is 2, as previous day it was 60, and so on.
+    Approach:
+            1. First approach is to scan backwards on each day and count the number of days stock was lesser than given day.
+               As soon we hit a bigger price, we stop. This solution is inefficient and would be O(n^2).
+            2. Solving it using stack.
+                1. Span of day1 is always 1. Put it on stack.
+                2. From day 2, We repeat next steps for all the remaining days
+                3. If price of the stock on the day on top of stack is less than price of stack on current day, pop it.
+                4. If price of the stock on the day on top of stack is greater than price of stock on current day, 
+                   calculate the span ( span = currentDay - day on top of stack)
+                5. Push the current day index on the stack.
+                6. Although it looks like it more than O(n), however each day is pushed on stack once and removed once, 
+                   Hence its O(n)

stack_problems/stock_span_problem.cpp

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+/*
+ * Stock span Problem
+ *
+ *   Given :   We have series of n daily price quotes for a stock. 
+ *
+ *   Required: We need to calculate **span** of stock's price for all n days.
+ *             Span for ith day is defined as **maximum** number of consecutive days,
+ *             for which the price of the stock was less than or equal to ith day.
+ *
+ *   Example:  For stock quotes {100, 60, 70, 65, 80, 85} span will be {1, 1, 2, 1, 4, 5}.
+ *             Span for day 1 is always 1, now for day 2 stock is at 60, and there is no day befor it when stock was less than 60.
+ *             So span remains 1. For day 3, the stock is priced at 70, so its span is 2, as previous day it was 60, and so on.
+ *   Refer readme of for approaches.
+ */  
+
+#include <iostream>
+#include <stack.h>
+
+void stock_spans( int *prices, int *spans, int days )
+{
+    algo::Stack<int> st;
+
+    //pushing day one to stack ( days are 0 indexed )
+    st.push(0);
+
+    //span for day 1
+    spans[0] = 1;
+
+    //span for rest of the days
+    for ( int i = 1; i < days; ++i )
+    {
+        // Pop till the day stock price was greater than today and stack is not emptuy.
+        while ( !st.empty() &&  prices[i] >= prices[st.top()] )
+        {
+            st.pop();
+        }
+        
+        /*
+         * If stack has emptied, then price[i] is the largest of price we have seen
+         * so far, else price[i] is greater than price on the day which is on top
+         * of stack.
+         */
+        spans[i] = (st.empty()) ? ( i + 1 ) : (i - st.top()) ;
+
+        //push today on stack.
+        st.push(i);
+    }
+}
+
+
+void printPricesSpans( int *prices, int *spans, int days )
+{
+    for ( int i = 0; i < days; ++i )
+    {
+        std::cout << "Day: " << (i+1)
+                  << " Price: " << prices[i]
+                  << " Span: " << spans[i]
+                  << std::endl;
+    }
+}
+
+
+int main()
+{
+    int prices[]= { 100, 60, 70, 65, 80, 85 }; 
+    int *spans = new int[6];
+    stock_spans(prices, spans, 6);
+    printPricesSpans( prices, spans, 6 );
+    return 0;
+}