LeetCode 11 – Container with most water

Posted on December 25, 2013

Last updated on December 25, 2013

Solution to LeetCode Container with most water problem.

The naive version would solve this in \(O(n^2)\) time. It can be improved as follows to linear time:

Keep two indices \(i\) and \(j\), that start on both ends of the array to process, height. Our final goal is to find \(i\),\(j\), such that \[ A(i,j) = (j-i) \times \min(\mathrm{height}(i), \mathrm{height}(j)) \] is maximized. At every step, if height[i] < height[j], move i up by one. Otherwise move j down by one, until the two pointers meet. We now show that this gives the maximal answer. Suppose we just computed \(A(i,j)\), and height[i] < height[j]. If we increase \(i\) by one, we will ignore all the values \(A(i,k)\) for \(i \leq k \lt j\). We now show that \(A(i,j) \gt A(i,k)\). Since height[i] < height[j], \(A(i,k)\) is bounded by the height[i] term, and since \(k < j\), it follows that \(A(i,j) \gt A(i,k)\), and we can safely ignore all of the \(A(i,k)\) and move on to inspecting \(A(i+1,j)\). The case with height[i] > height[j] is symmetric.

public int maxArea(int[] height) {
  int maxA = 0;
  int i = 0;
  int j = height.length-1;

  while(i < j){
    int curArea = (j-i) * (Math.min(height[i], height[j]));
    maxA = Math.max(curArea,maxA);
    if(height[i] < height[j]) i++;
    else j--;

  return maxA;
Container with most water
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