Given an NxN matrix of positive and negative integers, write code to find the submatrix with the largest possible sum. // "static void main" must be defined in a public class. public class Main { public static void main(String[] args) { System.o…

18.12 Given an NxN matrix of positive and negative integers, write code to find the submatrix with the largest possible sum. 这道求和最大的子矩阵,跟LeetCode上的Maximum Size Subarray Sum Equals k和Maximum Subarray很类似.这道题不建议使用brute force的方法,因为实在是不高效,我们需要借鉴上面LeetCode…

Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. Note:Given m satisfies the following const…

Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. Note: Given m satisfies the following cons…

17.8 You are given an array of integers (both positive and negative). Find the contiguous sequence with the largest sum. Return the sum. LeetCode上的原题,请参见我之前的博客Maximum Subarray. 解法一: int get_max_sum(vector<int> nums) { int res = INT_MIN, sum = INT_MI…

Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. Note: Given m satisfies the following cons…

做了Zenefits的OA,比面经里的简单多了..害我担心好久 阴险的Baidu啊,完全没想到用二分,一开始感觉要用DP,类似于极小极大值的做法. 然后看了答案也写了他妈好久. 思路是再不看M的情况下,最终结果的取值范围是[最大的元素,所有元素之和],然后就是用二分在这个范围里找. 注意二分的取舍要配合取舍方程,我用的方程是,保证子集的和都小于等于二分测试的那个二分M,然后集合数小于要求的m.一旦子集数量超过m,说明我们测试的二分M太小了,弄大点,才能让每个子集多放点元素,从而使得总子集数不超过…

Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. Note:If n is the length of array, assume t…

We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve? Note that our partition must use every number in A, and that scores are not…

题意大概就是,给定一个包含非负整数的序列nums以及一个整数m,要求把序列nums分成m份,并且要让这m个子序列各自的和的最大值最小(minimize the largest sum among these m subarrays). Note:If n is the length of array, assume the following constraints are satisfied: 1 ≤ n ≤ 10001 ≤ m ≤ min(50, n) Examples: Input:num…

Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. Examples: Input: nums = [7,2,5,10,8] m = 2…

Given a list of integers, write a function that returns the largest sum of non-adjacent numbers. Numbers can be 0 or negative. For example, [2, 4, 6, 2, 5] should return 13, since we pick 2, 6, and 5. [5, 1, 1, 5] should return 10, since we pick 5 an…

［抄题］: Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. Note:If n is the length of array, as…

Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these m subarrays. Note:Given m satisfies the following const…

We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve? Note that our partition must use every number in A, and that scores are not…

题目如下: Given an array which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays. Write an algorithm to minimize the largest sum among these msubarrays. Note:If n is the length of array, ass…

[LeetCode]813. Largest Sum of Averages 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/largest-sum-of-averages/description/ 题目描述: We partition a row of numbers A into at most…

When I finished reading this problem,I thought I could solve it by scanning every single subarray in the array,and the time complexity is cubic.Every subarray could be the eventual one whose sum is the largest,so I did make a conclusion that the best…

We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve? Note that our partition must use every number in A, and that scores are not…

We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve? Note that our partition must use every number in A, and that scores are not…

""" We partition a row of numbers A into at most K adjacent (non-empty) groups, then our score is the sum of the average of each group. What is the largest score we can achieve? Note that our partition must use every number in A, and that s…

对于一个数组中的数进行分组,取每个组里的平均值进行加和的. 使用动态规划,其中dp[i][k]表示以i为结尾的有k个分组的,那么递推式为: dp[i][k]=dp[j][k-1]+(sum[i]-sum[j])/(i-j)的,那么当k=1的时候就初始化为组内的平均值的,其中j的初始化为k-2,因为是从0为起始点的. class Solution { public: double largestSumOfAverages(vector<int>& A, int K) { int n=A.…

思路: dp. 实现: class Solution { public: int splitArray(vector<int>& nums, int m) { int n = nums.size(); vector<, ); ; i <= n; i++) sum[i] = sum[i - ] + nums[i - ]; vector<vector<, vector<, INT_MAX)); ; i <= n; i++) dp[i][] = sum[i…

给定一个非负整数数组和一个整数 m,你需要将这个数组分成 m 个非空的连续子数组.设计一个算法使得这 m 个子数组各自和的最大值最小.注意:数组长度 n 满足以下条件:    1 ≤ n ≤ 1000    1 ≤ m ≤ min(50, n)示例:输入:nums = [7,2,5,10,8]m = 2输出:18解释:一共有四种方法将nums分割为2个子数组.其中最好的方式是将其分为[7,2,5] 和 [10,8],因为此时这两个子数组各自的和的最大值为18,在所有情况中最小.详见:https:…

题目如下: 解题思路:求最值的题目优先考虑是否可以用动态规划.记dp[i][j]表示在数组A的第j个元素后面加上第i+1 (i从0开始计数)个分隔符后可以得到的最大平均值,那么可以得到递归关系式: dp[i][j] = max(dp[i][j],dp[i-1][k]+float(sum(A[k+1:j+1]))/float(j-k)) . 代码如下: class Solution(object): def largestSumOfAverages(self, A, K): ""&quo…

2018-07-12 23:21:53 问题描述: 问题求解: dp[i][j] : 以ai结尾的分j个部分得到的最大值 dp[i][j] = max{dp[k][j - 1] + (ak+1 + ... + ai) / (i - k)} k = [j - 2, i - 1] public double largestSumOfAverages(int[] A, int K) { double[][] dp = new double[A.length][K + 1]; int curSum =…

2019-10-14 22:13:18 问题描述: 问题求解: 解法一:动态规划 这种数组划分的题目基本都可以使用dp来解决,核心的思路就是先维护低的划分,再在中间找分割点加入新的划分. public int splitArray(int[] nums, int m) { int n = nums.length; long[][] dp = new long[m + 1][n]; long[] presum = new long[n]; presum[0] = nums[0]; for (int…

Chapter 1. Arrays and Strings 1.1 Unique Characters of a String 1.2 Reverse String 1.3 Permutation String 1.4 Replace Spaces 1.5 Compress String 1.6 Rotate Image 1.7 Set Matrix Zeroes 1.8 String Rotation Chapter 2. Linked Lists 2.1 Remove Duplicates…

given a 2-d matrix with 0 or 1 values largest square of all 1's dynamic programming, dp[i][j] = 1 + min{dp[i-1][j], dp[i][j-1], dp[i-1][j-1]} if m[i][j] == 1; otherwise a[i][j]=0; largest submatrix of all 1's https://leetcode.com/problems/maximal-rec…

Given a sequence of K integers { N1, N2, ..., NK }. A continuous subsequence is defined to be { Ni, Ni+1, ..., Nj } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For examp…