Watering Plants.cpp

/* Leet Code */
/* Title - Watering Plants */
/* Created By - Akash Modak */
/* Date - 05/05/2023 */

// You want to water n plants in your garden with a watering can. The plants are arranged in a row and are labeled from 0 to n - 1 from left to right where the ith plant is located at x = i. There is a river at x = -1 that you can refill your watering can at.

// Each plant needs a specific amount of water. You will water the plants in the following way:

// Water the plants in order from left to right.
// After watering the current plant, if you do not have enough water to completely water the next plant, return to the river to fully refill the watering can.
// You cannot refill the watering can early.
// You are initially at the river (i.e., x = -1). It takes one step to move one unit on the x-axis.

// Given a 0-indexed integer array plants of n integers, where plants[i] is the amount of water the ith plant needs, and an integer capacity representing the watering can capacity, return the number of steps needed to water all the plants.

 

// Example 1:

// Input: plants = [2,2,3,3], capacity = 5
// Output: 14
// Explanation: Start at the river with a full watering can:
// - Walk to plant 0 (1 step) and water it. Watering can has 3 units of water.
// - Walk to plant 1 (1 step) and water it. Watering can has 1 unit of water.
// - Since you cannot completely water plant 2, walk back to the river to refill (2 steps).
// - Walk to plant 2 (3 steps) and water it. Watering can has 2 units of water.
// - Since you cannot completely water plant 3, walk back to the river to refill (3 steps).
// - Walk to plant 3 (4 steps) and water it.
// Steps needed = 1 + 1 + 2 + 3 + 3 + 4 = 14.
// Example 2:

// Input: plants = [1,1,1,4,2,3], capacity = 4
// Output: 30
// Explanation: Start at the river with a full watering can:
// - Water plants 0, 1, and 2 (3 steps). Return to river (3 steps).
// - Water plant 3 (4 steps). Return to river (4 steps).
// - Water plant 4 (5 steps). Return to river (5 steps).
// - Water plant 5 (6 steps).
// Steps needed = 3 + 3 + 4 + 4 + 5 + 5 + 6 = 30.
// Example 3:

// Input: plants = [7,7,7,7,7,7,7], capacity = 8
// Output: 49
// Explanation: You have to refill before watering each plant.
// Steps needed = 1 + 1 + 2 + 2 + 3 + 3 + 4 + 4 + 5 + 5 + 6 + 6 + 7 = 49.

class Solution {
public:
    int wateringPlants(vector<int>& plants, int capacity) {
        int total=0,temp=capacity;
        for(int i=0;i<plants.size();i++) {
            if(temp<plants[i]) {
                total+=(2*i);
                temp=capacity;
            }
            total+=1;
            temp-=plants[i];
        }
        return total;
    }
};