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##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/math-fundamentals/basic-algebra/container-implementation

░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░░█████████████████████████████████████████████████████████████████████████████████████

#####   ######  #####    ###    #   #  ###  #   #  ######
##  ##  ##      ##  ##  ## ##   #   #   #   #   #  ##
#####   ####    #####   #   #   #   #   #   #   #  ####
##  #   ##      ##      ## ##    # #    #    # #   ##
##   #  ######  ##       ###      #    ###    #    ######

$ curl repovive.com/roadmaps/math-fundamentals/basic-algebra/container-implementation

Repovive

Container - Implementation - Basic Algebra | Math Fundamentals | Repovive

Repovive.

Math Fundamentals18 sections · 842 units13h total

Open in Course

Container - Implementation

In pseudocode

Here's the solution:


function maxArea(height)
    left := 0
    right := length of height - 1
    maxArea := 0
    while left < right
        width := right - left
        currentHeight := min(height[left], height[right])
        currentArea := width × currentHeight
        maxArea := max(maxArea, currentArea)
        if height[left] < height[right] then
            left := left + 1
        else
            right := right - 1
    return maxArea

This runs in $O(n)$ time with $O(1)$ space. Each pointer moves at most $n$ times.