This content originally appeared on DEV Community and was authored by Flame Chan
122. Best Time to Buy and Sell Stock II
You are given an integer array prices where prices[i] is the price of a given stock on the ith day.
On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.
Find and return the maximum profit you can achieve.
Example 1:
Input: prices = [7,1,5,3,6,4]
Output: 7
Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
Total profit is 4 + 3 = 7.
Example 2:
Input: prices = [1,2,3,4,5]
Output: 4
Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Total profit is 4.
Example 3:
Input: prices = [7,6,4,3,1]
Output: 0
Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.
Constraints:
1 <= prices.length <= 3 * 104
0 <= prices[i] <= 104
Original Page
Wrong Code
public int maxProfit(int[] prices) {
int profit = 0;
int buy = Integer.MAX_VALUE;
int sum = 0;
int peek = 0;
for(int i=0; i<prices.length; i++){
int num = prices[i];
if(num > buy && num > peek){
profit = num - buy;
peek = num;
}
else if((num>buy && num<peek) || num < buy){
sum += profit;
profit = 0;
buy = num;
peek = num;
}
}
return sum+profit;
}
I init buy to int MAX_VALUE and forget to update this may lead to some error.
fix this and truncate the codes
Fine Code
public int maxProfit(int[] prices) {
if(prices.length <1){
return 0;
}
int profit = 0;
int buy = prices[0];
int sum = 0;
int peek = prices[0];
for(int i=0; i<prices.length; i++){
int num = prices[i];
if( num > peek){
profit = num - buy;
peek = num;
}
else if(num<peek){
sum += profit;
profit = 0;
buy = num;
peek = num;
}
}
sum += profit;
return sum;
}
1005. Maximize Sum Of Array After K Negations
Given an integer array nums and an integer k, modify the array in the following way:
choose an index i and replace nums[i] with -nums[i].
You should apply this process exactly k times. You may choose the same index i multiple times.
Return the largest possible sum of the array after modifying it in this way.
Example 1:
Input: nums = [4,2,3], k = 1
Output: 5
Explanation: Choose index 1 and nums becomes [4,-2,3].
Example 2:
Input: nums = [3,-1,0,2], k = 3
Output: 6
Explanation: Choose indices (1, 2, 2) and nums becomes [3,1,0,2].
Example 3:
Input: nums = [2,-3,-1,5,-4], k = 2
Output: 13
Explanation: Choose indices (1, 4) and nums becomes [2,3,-1,5,4].
Constraints:
1 <= nums.length <= 10^4
-100 <= nums[i] <= 100
1 <= k <= 10^4
Original Page
public int largestSumAfterKNegations(int[] nums, int k) {
Arrays.sort(nums);
int change = nums[nums.length-1] >=0?0:nums.length-1;
int sum = 0;
for(int i=0; i<nums.length; i++){
if(nums[i] < 0 && k>0){
sum -= nums[i];
k--;
}else{
sum += nums[i];
}
// find the cross over
if(i>0 && nums[i-1]<=0 && nums[i]>0){
if(-nums[i-1]>nums[i]){
change = i;
}else{
change = i-1;
}
}
}
// k>nums.length so we need to run out of these k
if(k>0){
if(k%2!=0){
//find the min abs value
sum -= 2*Math.abs(nums[change]);
}
}
return sum;
}
55. Jump Game
You are given an integer array nums. You are initially positioned at the array's first index, and each element in the array represents your maximum jump length at that position.
Return true if you can reach the last index, or false otherwise.
Example 1:
Input: nums = [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:
Input: nums = [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum jump length is 0, which makes it impossible to reach the last index.
Constraints:
1 <= nums.length <= 10^4
0 <= nums[i] <= 10^5
Wrong Code
public boolean canJump(int[] nums) {
//find whether achive the last element so that we only can see whether can reach the second last element
for(int i=0; i<nums.length-1;){
int size = nums[i];
int next = 0;
int nextVal = 0;
if(size == 0){
return false;
}
if(i+size >= nums.length){
return true;
}
//find the max steps among the current step
for(int j=0; j<=size; j++){
// calculate max for both index and value
if(i+j+nums[i+j]>next){
next = i+j+nums[i+j];
}
}
i = next;
}
return true;
}
Wrong Code 2
public boolean canJump(int[] nums) {
if(nums.length==1){
return true;
}
if(nums[0] == 0){
return false;
}
for(int i=0; i<nums.length-1; i++){
if(i+nums[i]>=nums.length-1){
return true;
}
}
return false;
}
public boolean canJump(int[] nums) {
if(nums.length==1){
return true;
}
int range = 0;
for(int i=0; i<=range; i++){
range = Math.max(range, i+nums[i]);
if(range >= nums.length-1){
return true;
}
}
return false;
}
45. Jump Game II
You are given a 0-indexed array of integers nums of length n. You are initially positioned at nums[0].
Each element nums[i] represents the maximum length of a forward jump from index i. In other words, if you are at nums[i], you can jump to any nums[i + j] where:
0 <= j <= nums[i] and
i + j < n
Return the minimum number of jumps to reach nums[n - 1]. The test cases are generated such that you can reach nums[n - 1].
Example 1:
Input: nums = [2,3,1,1,4]
Output: 2
Explanation: The minimum number of jumps to reach the last index is 2. Jump 1 step from index 0 to 1, then 3 steps to the last index.
Example 2:
Input: nums = [2,3,0,1,4]
Output: 2
Constraints:
1 <= nums.length <= 104
0 <= nums[i] <= 1000
It's guaranteed that you can reach nums[n - 1].
public int jump(int[] nums) {
if(nums.length == 1){
return 0;
}
int step = 0;
int range = 0;
int preRange = 0;
for(int i=0; i<nums.length-1; i++){
range = Math.max(range, i+nums[i]);
if(range >= nums.length-1){
step++;
break;
}
if(i == preRange){
preRange = range;
step++;
}
}
return step;
}
This content originally appeared on DEV Community and was authored by Flame Chan
Flame Chan | Sciencx (2024-07-03T11:32:28+00:00) LeetCode Day24 Greedy Algorithms Part 2. Retrieved from https://www.scien.cx/2024/07/03/leetcode-day24-greedy-algorithms-part-2/
Please log in to upload a file.
There are no updates yet.
Click the Upload button above to add an update.