Given an array walls of wall heights, calculate the maximum volume of water that will be trapped between two walls and the x-axis after a heavy downpour.
walls: number[]: An array of integersInput: walls = [1,4,2,3]Output: 6Explanation: Consider two walls (i=1 & i=3) with heights 4 and 3. The water is limited by the shorter wall with height 3, so the container holds 2 (distance) * 3 (shorter height) = 6 units. All other combination of walls result in smaller area.
Input: walls = [1,1]Output: 1Explanation: Consider two walls (i=0 & i=1) with heights 1 and 1. The water is limited by the shorter wall (both same in this case i.e. 1), so the container holds 1 (distance) * 1 (shorter height) = 1 unit
Input: walls = [1,0]Output: 0Explanation: Consider two walls (i=0 & i=1) with heights 1 and 0. The water is limited by the shorter wall with height 0, so the container holds 1 (distance) * 0 (shorter height) = 0 unit
walls.length <= 1000walls[i] <= 10,000Given an array walls of wall heights, calculate the maximum volume of water that will be trapped between two walls and the x-axis after a heavy downpour.
walls: number[]: An array of integersInput: walls = [1,4,2,3]Output: 6Explanation: Consider two walls (i=1 & i=3) with heights 4 and 3. The water is limited by the shorter wall with height 3, so the container holds 2 (distance) * 3 (shorter height) = 6 units. All other combination of walls result in smaller area.
Input: walls = [1,1]Output: 1Explanation: Consider two walls (i=0 & i=1) with heights 1 and 1. The water is limited by the shorter wall (both same in this case i.e. 1), so the container holds 1 (distance) * 1 (shorter height) = 1 unit
Input: walls = [1,0]Output: 0Explanation: Consider two walls (i=0 & i=1) with heights 1 and 0. The water is limited by the shorter wall with height 0, so the container holds 1 (distance) * 0 (shorter height) = 0 unit
walls.length <= 1000walls[i] <= 10,000console.log() statements will appear here.