455. Assign Cookies
Assume you are an awesome parent and want to give your children some cookies. But, you should give each child at most one cookie.
Each child i has a greed factor g[i], which is the minimum size of a cookie that the child will be content with; and each cookie j has a size s[j]. If s[j] >= g[i], we can assign the cookie j to the child i, and the child i will be content. Your goal is to maximize the number of your content children and output the maximum number.
Example 1:
Input: g = [1,2,3], s = [1,1]
Output: 1
Explanation: You have 3 children and 2 cookies. The greed factors of 3 children are 1, 2, 3.
And even though you have 2 cookies, since their size is both 1, you could only make the child whose greed factor is 1 content.
You need to output 1.
Example 2:
Input: g = [1,2], s = [1,2,3]
Output: 2
Explanation: You have 2 children and 3 cookies. The greed factors of 2 children are 1, 2.
You have 3 cookies and their sizes are big enough to gratify all of the children,
You need to output 2.
Constraints:
1 <= g.length <= 3 * 104
0 <= s.length <= 3 * 104
1 <= g[i], s[j] <= 231 - 1
public int findContentChildren(int[] g, int[] s) {
// avoid null pointer
if(g.length == 0 || s.length ==0){
return 0;
}
// 2 * nlogn
Arrays.sort(g);
Arrays.sort(s);
int i = 0;
int j = 0;
int count = 0;
while(i < g.length && j < s.length){
if(g[i] <= s[j]){
i++;
j++;
count++;
} else{
j++;
}
}
return count;
}
time : n`logn
Another version for loop
`
public int findContentChildren(int[] g, int[] s) {
// avoid null pointer
if(g.length == 0 || s.length ==0){
return 0;
}
// 2 * nlogn
Arrays.sort(g);
Arrays.sort(s);
int j = 0;
int count = 0;
for(int i=0; i<s.length && j<g.length; i++){
if(s[i] >= g[j]){
j++;
count++;
}
}
return count;
}
`
376. Wiggle Subsequence
A wiggle sequence is a sequence where the differences between successive numbers strictly alternate between positive and negative. The first difference (if one exists) may be either positive or negative. A sequence with one element and a sequence with two non-equal elements are trivially wiggle sequences.
For example, [1, 7, 4, 9, 2, 5] is a wiggle sequence because the differences (6, -3, 5, -7, 3) alternate between positive and negative.
In contrast, [1, 4, 7, 2, 5] and [1, 7, 4, 5, 5] are not wiggle sequences. The first is not because its first two differences are positive, and the second is not because its last difference is zero.
A subsequence is obtained by deleting some elements (possibly zero) from the original sequence, leaving the remaining elements in their original order.
Given an integer array nums, return the length of the longest wiggle subsequence of nums.
Example 1:
Input: nums = [1,7,4,9,2,5]
Output: 6
Explanation: The entire sequence is a wiggle sequence with differences (6, -3, 5, -7, 3).
Example 2:
Input: nums = [1,17,5,10,13,15,10,5,16,8]
Output: 7
Explanation: There are several subsequences that achieve this length.
One is [1, 17, 10, 13, 10, 16, 8] with differences (16, -7, 3, -3, 6, -8).
Example 3:
Input: nums = [1,2,3,4,5,6,7,8,9]
Output: 2
Constraints:
1 <= nums.length <= 1000
0 <= nums[i] <= 1000
Follow up: Could you solve this in O(n) time?
`
public int wiggleMaxLength(int[] nums) {
if(nums.length == 0){
return 0;
}
int count = 1;
int preFlag = 0;
int pre = nums[0];
for(int i=1; i<nums.length; i++){
if(nums[i]-nums[i-1] != 0){
int flag = (nums[i]-nums[i-1])/Math.abs(nums[i]-nums[i-1]);
if(flag == -preFlag || preFlag == 0){
preFlag = flag;
count++;
}
}
}
return count;
}
`
53. Maximum Subarray
Given an integer array nums, find the
subarray
with the largest sum, and return its sum.
Example 1:
Input: nums = [-2,1,-3,4,-1,2,1,-5,4]
Output: 6
Explanation: The subarray [4,-1,2,1] has the largest sum 6.
Example 2:
Input: nums = [1]
Output: 1
Explanation: The subarray [1] has the largest sum 1.
Example 3:
Input: nums = [5,4,-1,7,8]
Output: 23
Explanation: The subarray [5,4,-1,7,8] has the largest sum 23.
Constraints:
1 <= nums.length <= 105
-104 <= nums[i] <= 104
Follow up: If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.
`
public int maxSubArray(int[] nums) {
if(nums.length == 0){
return 0;
}
int max = Integer.MIN_VALUE;
int sum = Integer.MIN_VALUE;
for(int i=0; i
if(nums[i] > 0){
if(sum < 0){
sum = nums[i];
}else{
sum += nums[i];
}
max = Math.max(max, sum);
}else{
if(sum<0){
sum = nums[i];
}else{
sum += nums[i];
}
max = Math.max(max, sum);
}
}
return max;
}
`
improve code
`
public int maxSubArray(int[] nums) {
if(nums.length == 0){
return 0;
}
int max = Integer.MIN_VALUE;
int sum = Integer.MIN_VALUE;
for(int i=0; i<nums.length; i++){
if(sum < 0){
sum = nums[i];
}else{
sum += nums[i];
}
max = Math.max(max, sum);
}
return max;
}
`
Another way for greedy
`
public int maxSubArray(int[] nums) {
if(nums.length == 0){
return 0;
}
int max = Integer.MIN_VALUE;
// int sum = Integer.MIN_VALUE;
int sum = 0;
for(int i=0; i<nums.length; i++){
sum+= nums[i];
if(max < sum){
max = sum;
}
if(sum <0){
sum = 0;
}
// if(sum < 0){
// sum = nums[i];
// }else{
// sum += nums[i];
// }
// max = Math.max(max, sum);
}
return max;
}
`
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