• Expected Output: 12
• Justification: The maximum rectangular area is formed by the bars of heights 5, 4, 5, which is equal to 4x3=12. This rectangle spans from the third to the sixth bar.

Solution

To solve this problem, the key approach is to utilize a stack to keep track of bars that are potentially part of the largest rectangle. We traverse each bar in the histogram and use the stack to maintain bars in ascending order of height. When we encounter a bar that is shorter than the bar at the top of the stack, it indicates the end of a potentially largest rectangle involving the top-of-stack bar. By calculating the area of the rectangle with the stack's top bar as the smallest bar and comparing it with the maximum area found so far, we ensure that we find the largest rectangle.

This method is effective because it allows us to dynamically adjust to the changing potential for the largest rectangle as we iterate through the bars. It combines the efficiency of a single pass through the data with the ability to backtrack and calculate areas as soon as we know a rectangle's bounds. This blend of forward-looking and immediate calculation ensures we efficiently find the largest possible rectangle.

Step-by-step Algorithm

• Initialize an empty stack to keep track of the indices of bars.
• Iterate through each bar in the histogram.
  • While the stack is not empty and the current bar's height is less than the height of the bar at the top of the stack:
    • Pop the top of the stack. Let this be topIndex.
    • Calculate the area of the rectangle using the height of the popped bar. If the stack is empty, the width is the current index i; otherwise, it's i - stack.peek() - 1.
    • Update the maximum area if necessary.
  • Push the current index onto the stack.
• Return the maximum area found.

Algorithm Walkthrough

Let's consider the input [6,2,5,4,5,1,6]:

• Step 1: Start with an empty stack and maxArea = 0.
• Step 2: Iterate through each bar's height:
  • At index 0 (height=6):
    • Push 0 onto the stack.
    • maxArea remains 0.
  • At index 1 (height=2):
    • Since 6 > 2, pop index 0 from the stack. Calculate area: 6 * (1 - 0) = 6. Update maxArea to 6.
    • Push index 1 onto the stack.
  • At index 2 (height=5):
    • Push index 2 onto the stack since 2 < 5. maxArea remains 6.
  • At index 3 (height=4):
    • Since 5 > 4, pop index 2 from the stack. Calculate area: 5 * (3 - 1 - 1) = 5. maxArea remains same.
    • Push index 3 onto the stack.
  • At index 4 (height=5):
    • Push index 4 onto the stack since 4 < 5. maxArea remains 6.
  • At index 5 (height=1):
    • Since 5 > 1, pop index 4 from the stack. Calculate area: 5 * (5 - 3 - 1) = 5. maxArea remains 6.
    • Since 4 > 1, pop index 3 from the stack. Calculate area: 4 * (5 - 1 - 1) = 12. Update maxArea to 12.
    • Since 2 > 1, pop index 1 from the stack. Calculate area: 2 * (i) = 2 * (5) = 10. maxArea remains same 12.
    • Push index 5 onto the stack.
  • At index 6 (height=6):
    • Since 1 < 6, push index 6 onto the stack. maxArea remains 12.
• Final maxArea is 12.

Code