Problem Statement:
There are n gas stations along a circular route, where the amount of gas at the ith station is gas[i].
You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the ith station to its next (i + 1)th station. You begin the journey with an empty tank at one of the gas stations.
Given two integer arrays gas and cost, return the starting gas station’s index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique.
Example 1:
Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
Output: 3
Explanation: Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4 Travel to station 4. Your tank = 4 – 1 + 5 = 8 Travel to station 0. Your tank = 8 – 2 + 1 = 7 Travel to station 1. Your tank = 7 – 3 + 2 = 6 Travel to station 2. Your tank = 6 – 4 + 3 = 5 Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3. Therefore, return 3 as the starting index.
Example 2:
Input: gas = [2,3,4], cost = [3,4,3]
Output: -1
Explanation: You can’t start at station 0 or 1, as there is not enough gas to travel to the next station. Let’s start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4 Travel to station 0. Your tank = 4 – 3 + 2 = 3 Travel to station 1. Your tank = 3 – 3 + 3 = 3 You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3. Therefore, you can’t travel around the circuit once no matter where you start.
Constraints
n == gas.length == cost.length1 <= n <= 1050 <= gas[i], cost[i] <= 104- The input is generated such that the answer is unique.
Approach:
- Calculate the net gain
(gas[i] - cost[i])at each station. - Keep two sums:
totalGain: total net gain for the whole circuit.currGain: running sum for the current starting point.
- If
currGainbecomes negative, it means you cannot start from the current segment. So, resetcurrGain = 0and set the next station(i+1)as the new candidate starting point. - After the loop:
- If
totalGain < 0, return-1(not possible to complete circuit). - Otherwise, return ans (the valid starting station index).
- If
Time & Space Complexity:
Time Complexity: O(n)
Space Complexity: O(1)
Dry Run
Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
Step 0: Start Function: canCompleteCircuit(gas, cost) gas = [1,2,3,4,5], cost = [3,4,5,1,2] currGain = 0, totalGain = 0, ans = 0 Iteration (i = 0): - gain = gas[0] - cost[0] = 1 - 3 = -2 - currGain = 0 + (-2) = -2 - totalGain = 0 + (-2) = -2 - currGain < 0 → true → ans = 0 + 1 = 1, currGain = 0 Iteration (i = 1): - gain = gas[1] - cost[1] = 2 - 4 = -2 - currGain = 0 + (-2) = -2 - totalGain = -2 + (-2) = -4 - currGain < 0 → true → ans = 1 + 1 = 2, currGain = 0 Iteration (i = 2): - gain = gas[2] - cost[2] = 3 - 5 = -2 - currGain = 0 + (-2) = -2 - totalGain = -4 + (-2) = -6 - currGain < 0 → true → ans = 2 + 1 = 3, currGain = 0 Iteration (i = 3): - gain = gas[3] - cost[3] = 4 - 1 = 3 - currGain = 0 + 3 = 3 - totalGain = -6 + 3 = -3 - currGain < 0 → false → ans stays = 3 Iteration (i = 4): - gain = gas[4] - cost[4] = 5 - 2 = 3 - currGain = 3 + 3 = 6 - totalGain = -3 + 3 = 0 - currGain < 0 → false → ans stays = 3 Step 3: End - totalGain = 0 → not < 0 - return ans = 3
Output: 3
var canCompleteCircuit = function (gas, cost) {
let currGain = 0;
let totalGain = 0;
let ans = 0;
for (let i = 0; i < gas.length; i++) {
let gain = gas[i] - cost[i];
currGain += gain;
totalGain += gain;
if (currGain < 0) {
ans = i + 1;
currGain = 0;
}
}
return totalGain < 0 ? -1 : ans;
};
def canCompleteCircuit(gas, cost):
currGain = 0
totalGain = 0
ans = 0
for i in range(len(gas)):
gain = gas[i] - cost[i]
currGain += gain
totalGain += gain
if currGain < 0:
ans = i + 1
currGain = 0
return -1 if totalGain < 0 else ans
public class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
long currGain = 0;
long totalGain = 0;
int ans = 0;
int n = gas.length;
for (int i = 0; i < n; i++) {
long gain = (long)gas[i] - (long)cost[i];
currGain += gain;
totalGain += gain;
if (currGain < 0) {
ans = i + 1;
currGain = 0;
}
}
return (totalGain < 0) ? -1 : ans;
}
}
#include <vector>
#include <cstdint>
using namespace std;
int canCompleteCircuit(const vector& gas, const vector& cost) {
long long currGain = 0;
long long totalGain = 0;
int ans = 0;
int n = gas.size();
for (int i = 0; i < n; ++i) {
long long gain = (long long)gas[i] - (long long)cost[i];
currGain += gain;
totalGain += gain;
if (currGain < 0) {
ans = i + 1;
currGain = 0;
}
}
return (totalGain < 0) ? -1 : ans;
}
#include <stdio.h>
int canCompleteCircuit(int *gas, int *cost, int n) {
long long currGain = 0;
long long totalGain = 0;
int ans = 0;
for (int i = 0; i < n; ++i) {
long long gain = (long long)gas[i] - (long long)cost[i];
currGain += gain;
totalGain += gain;
if (currGain < 0) {
ans = i + 1;
currGain = 0;
}
}
return (totalGain < 0) ? -1 : ans;
}
using System;
public class Solution {
public int CanCompleteCircuit(int[] gas, int[] cost) {
long currGain = 0;
long totalGain = 0;
int ans = 0;
int n = gas.Length;
for (int i = 0; i < n; i++) {
long gain = (long)gas[i] - (long)cost[i];
currGain += gain;
totalGain += gain;
if (currGain < 0) {
ans = i + 1;
currGain = 0;
}
}
return (totalGain < 0) ? -1 : ans;
}
}
1 Comment
Soi cầu lô rơi từ giải Đặc biệt
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