## Sunday, October 2, 2016

### LeetCode 410 - Split Array Largest Sum

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 constraint: 1 ≤ m ≤ length(nums) ≤ 14,000.
Examples:
```Input:
nums = [1,2,3,4,5]
m = 2

Output:
9

Explanation:
There are four ways to split nums into two subarrays.
The best way is to split it into [1,2,3] and [4,5],
where the largest sum among the two subarrays is only 9.```
X. Bisection method
https://discuss.leetcode.com/topic/61315/java-easy-binary-search-solution-8ms
1. Given a result, it is easy to test whether it is valid or not.
2. The max of the result is the sum of the input nums.
3. The min of the result is the max num of the input nums.
Given the 3 conditions above we can do a binary search. (need to deal with overflow)
``````    public int splitArray(int[] nums, int m) {
long max = 0, min = 0;
for(int num: nums) {
max += num;
min = Math.max(num, min);
}
return (int) binarySearch(nums, min, max, m);
}
private long binarySearch(int[] nums, long min, long max, int m) {
if(min == max) return min;
long mid = (max - min) / 2 + min;
if(isValid(nums, mid, m)) {
if(mid == min || !isValid(nums, mid - 1, m)) return mid;
else return binarySearch(nums, min, mid - 1, m);
} else return binarySearch(nums, mid + 1, max, m);
}``````
https://discuss.leetcode.com/topic/61324/clear-explanation-8ms-binary-search-java
1. The answer is between maximum value of input array numbers and sum of those numbers.
2. Use binary search to approach the correct answer. We have `l = max number of array; r = sum of all numbers in the array;`Every time we do `mid = (l + r) / 2;`
3. Use greedy to narrow down left and right boundaries in binary search.
3.1 Cut the array from left.
3.2 Try our best to make sure that the sum of numbers between each two cuts (inclusive) is large enough but still less than `mid`.
3.3 We'll end up with two results: either we can divide the array into more than m subarrays or we cannot.
``````    public int splitArray(int[] nums, int m) {
long sum = 0;
int max = 0;
for(int num: nums){
max = Math.max(max, num);
sum += num;
}
return (int)binary(nums, m, sum, max);
}

private long binary(int[] nums, int m, long high, long low){
long mid = 0;
while(low < high){
mid = (high + low)/2;
if(valid(nums, m, mid)){
//System.out.println(mid);
high = mid;
}else{
low = mid + 1;
}
}
return high;
}

private boolean valid(int[] nums, int m, long max){
int cur = 0;
int count = 1;
for(int num: nums){
cur += num;
if(cur > max){
cur = num;
count++;
if(count > m){
return false;
}
}
}
return true;
}``````
http://blog.csdn.net/mebiuw/article/details/52724293
/** * 这个数肯定介于最大的那一个单值和所有元素只和的中间 * */ public int splitArray(int[] nums, int m) { long sum = 0; int max = 0; for(int num: nums){ max = Math.max(max, num); sum += num; } return (int)binarySearch(nums, m, sum, max); } //二分查找 private long binarySearch(int[] nums, int m, long high, long low){ long mid = 0; while(low < high){ mid = (high + low)/2; //验证是否满足,也就是这么大的值有可能出现么 if(valid(nums, m, mid)){ high = mid; }else{ low = mid + 1; } } return high; } /** * 验证这个值是否合法 * */ private boolean valid(int[] nums, int m, long max){ int cur = 0; int count = 1; //是否有多余m个片段or区间，大于给定值的max的，如果有了，那么就不合法了，因为这样划分就不止m个，及max太小 for(int num: nums){ cur += num; if(cur > max){ cur = num; count++; if(count > m){ return false; } } } return true; }
二分枚举答案（Binary Search）
将数组nums拆分成m个子数组，每个子数组的和不小于sum(nums) / m，不大于sum(nums)
又因为数组nums中只包含非负整数，因此可以通过二分法在上下界内枚举答案。
时间复杂度O(n * log m)，其中n是数组nums的长度，m为数组nums的和
http://www.cnblogs.com/grandyang/p/5933787.html
https://all4win78.wordpress.com/2016/10/22/leetcode-410-split-array-largest-sum/
另一种就是利用sum是整数的性质进项binary search。UB (upper bound) 是所有原array的所有数字的和，LB (lower bound) 是UB/m以数组中最大值两者之间的较大者。然后根据binary search的方式去验证给定的array能否依据这个largest sum进行分割。需要注意的是，这里sum需要使用long而不是int，不然会有overflow的问题。世间复杂度是O(n log sum) <= O(n log n*Integer.MAX_VALUE) <= O(n (logn + c)) = O(n logn)，额外的空间复杂度为O(1)。
http://blog.csdn.net/mebiuw/article/details/52724293

http://blog.csdn.net/u014688145/article/details/69525838
如果直接按照上述的意思去写代码时，你会发现实操非常困难，所以必须对问题归简，这道题还可以这样理解，求最小的最大，关键在于最小，而不是最大！真正做约束的在于最小，针对每种划分情况，最大是每种划分的固有属性，所以完全可以不用考虑，否则增添理解题目的负担。那么最小的含义在于，对于`m=2`的情况，我们是尽可能的让左右两部分平衡，也就是说`sum(左数组) = sum(右数组)`，也就是平均分配。
先说说我的想法吧，但我的代码没有AC，后来发现，这种做法没法实现`m >= 3`的情况，太可惜了，但`m == 2`是完全适用的。
我的解法典型的给你`nums,m`想办法去求解`minMaxsum`，而大神的思路是假设我们已经在解空间里有了一系列`minMaxsums`，去搜索一个`minMaxsum`使得符合`m`，这让我非常震撼。
很简单，可能的minMaxsum有哪些，中间的哪些minMaxsum我们是不知道的，这是问题的关键！所以这个问题就假设最极端的两头情况：
• 当 m = 1 时，这种情况，minMaxsum = 整个数组之和。
• 当 m = 数组长度时，这种情况，minMaxsum = 数组中最大的那个元素。
如：[7,2,5,10,8] m = 1， 输出 minMaxsum = 32
而：[7,2,5,10,8] m = 5， 输出 minMaxsum = 10
这里让我对二分查找有了重新的认识：
• 下标不一定是数组的下标，求解问题的解空间同样可以当作下标，满足状态递增就行。
• 一定要符合状态空间中元素的有序性么？不一定，只要在搜索时，能够有约束条件排除左半或者右半即可。

搜索解空间，找对应的`m`，符合情况就输出，很好的一个思路。
public int splitArray(int[] nums, int m) { long sum = 0; int max = 0; for(int num: nums){ max = Math.max(max, num); sum += num; } return (int)binary(nums, m, sum, max); } private long binary(int[] nums, int m, long high, long low){ long mid = 0; while(low < high){ mid = (high + low)/2; if(valid(nums, m, mid)){ high = mid; }else{ low = mid + 1; } } return high; } private boolean valid(int[] nums, int m, long max){ int cur = 0; int count = 1; for(int num: nums){ cur += num; if(cur > max){ cur = num; count++; if(count > m){ return false; } } } return true; }

X. DP
http://www.cnblogs.com/grandyang/p/5933787.html
我们建立一个二维数组dp，其中dp[i][j]表示将数组中前j个数字分成i组所能得到的最小的各个子数组中最大值，初始化为整型最大值，如果无法分为i组，那么还是保持为整型最大值。为了能快速的算出子数组之和，我们还是要建立累计和数组，难点就是在于要求递推公式了。我们来分析，如果前j个数字要分成i组，那么i的范围是什么，由于只有j个数字，如果每个数字都是单独的一组，那么最多有j组；如果将整个数组看为一个整体，那么最少有1组，所以i的范围是[1, j]，所以我们要遍历这中间所有的情况，假如中间任意一个位置k，dp[i-1][k]表示数组中前k个数字分成i-1组所能得到的最小的各个子数组中最大值，而sums[j]-sums[k]就是后面的数字之和，我们取二者之间的较大值，然后和dp[i][j]原有值进行对比，更新dp[i][j]为二者之中的较小值，这样k在[1, j]的范围内扫过一遍，dp[i][j]就能更新到最小值，我们最终返回dp[m][n]即可，博主认为这道题所用的思想应该是之前那道题Reverse Pairs中解法二中总结的分割重现关系(Partition Recurrence Relation)，由此看来很多问题的本质都是一样，但是披上华丽的外衣，难免会让人有些眼花缭乱了
```    int splitArray(vector<int>& nums, int m) {
int n = nums.size();
vector<int> sums(n + 1, 0);
vector<vector<int>> dp(m + 1, vector<int>(n + 1, INT_MAX));
dp[0][0] = 0;
for (int i = 1; i <= n; ++i) {
sums[i] = sums[i - 1] + nums[i - 1];
}
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
for (int k = i - 1; k < j; ++k) {
int val = max(dp[i - 1][k], sums[j] - sums[k]);
dp[i][j] = min(dp[i][j], val);
}
}
}
return dp[m][n];
}```

DP的话实现起来稍微麻烦一点，需要保存一个2d的array，大小为n*m，每个位置（i，j）保存从数列位置i开始到数列尾端，分割成j份的largest sum。同时空间和时间复杂度在m比较大的情况下比较高，并不是很建议勇这种方法implement，但是可以作为一种思路。类似的题目可以参考Burst Balloons
https://discuss.leetcode.com/topic/66289/java-dp
``````public int splitArray(int[] nums, int m) {
int[][] memo = new int[nums.length][m + 1];
int[] sum = new int[nums.length];
sum[nums.length - 1] = nums[nums.length - 1];
for(int i = nums.length - 2; i >= 0; i--){
sum[i] = sum[i + 1] + nums[i];
}
return findSA(nums, 0, m, sum, memo);
}

public int findSA(int[] nums, int start, int m, int[] sums, int[][] memo){
if(m == 1) return sums[start];
if(memo[start][m] > 0)
return memo[start][m];
int min = Integer.MAX_VALUE, sum = 0;
for(int i = start; i <= nums.length - m; i++){
sum += nums[i];
min = Math.min(Math.max(sum, findSA(nums, i + 1, m - 1, sums, memo)), min);
}
memo[start][m] = min;
return memo[start][m];
}``````

time complexity of naive dp is O(n2 * m)
``````    public int splitArray(int[] nums, int K) {
int[] s = new int[nums.length];
s[0] = nums[0];
for (int i = 1; i < nums.length; i++) {
s[i] = nums[i] + s[i - 1];
}

for (int k = 2; k <= K; k++) {
for (int i = nums.length - 1; i >= k - 1; i--) {
int min = Integer.MAX_VALUE;
int left = nums[i];
for (int p = i - 1; p >= k - 2; p--) {
min = Math.min(min, Math.max(s[p], left));
left += nums[p];
if (left >= min) {
break;
}
}
s[i] = min;
}
}

return s[nums.length - 1];
}``````
https://discuss.leetcode.com/topic/61405/dp-java
This is obviously not as good as the binary search solutions; but it did pass OJ.
`dp[s,j]` is the solution for splitting subarray `n[j]...n[L-1]` into `s` parts.
`dp[s+1,i] = min{ max(dp[s,j], n[i]+...+n[j-1]) }, i+1 <= j <= L-s`
This solution does not take advantage of the fact that the numbers are non-negative (except to break the inner loop early). That is a loss. (On the other hand, it can be used for the problem containing arbitrary numbers)
``````public int splitArray(int[] nums, int m)
{
int L = nums.length;
int[] S = new int[L+1];
S[0]=0;
for(int i=0; i<L; i++)
S[i+1] = S[i]+nums[i];

int[] dp = new int[L];
for(int i=0; i<L; i++)
dp[i] = S[L]-S[i];

for(int s=1; s<m; s++)
{
for(int i=0; i<L-s; i++)
{
dp[i]=Integer.MAX_VALUE;
for(int j=i+1; j<=L-s; j++)
{
int t = Math.max(dp[j], S[j]-S[i]);
if(t<=dp[i])
dp[i]=t;
else
break;
}
}
}

return dp[0];
}``````
Elegant solution. Based on your solution I did some optimization, to short cut some un-necessary search:
``````public static int splitArray(int[] nums, int m) {
int[] dp = new int[nums.length];

for(int i = nums.length-1; i>=0; i--)
dp[i] = i== nums.length -1 ? nums[i] : dp[i+1] + nums[i];
for(int im = 2; im <= m; im ++) {
int maxPart = nums.length + 1 - im;
for(int i=0; i<maxPart; i++) {
dp[i] = Integer.MAX_VALUE;
int leftSum = 0;
for(int p=i; p<maxPart; p++) {
leftSum += nums[p];
if(leftSum > dp[i])
break;  // There's no more better soluiton, stop the search.
int val = Math.max(leftSum, dp[p+1]);
if(val < dp[i])
dp[i] = val;
}
if(im == m)  // The last round, get first one is enough
break;
}
}
return dp[0];
}``````
You can add the sum on the run, you can compare the sum only since sum is accending and dp[j+1] is descending. Unlike the binary search solution, O(nlogk) (k: related to the range of num), this one is not related to the num range, O(n^2*m), I think it could do better for certain test cases, such as [700000000,200000000,500000000,1000000000,800000000], 2
``````public int splitArray(int[] nums, int m) {
int n = nums.length;
int[] dp = new int[n+1];
for (int i = n-1; i >= m-1; --i) {
dp[i] = dp[i+1] + nums[i];
}
for (int k = 2; k <= m; ++k) {
for (int i = m-k; i <= n-k; ++i) {
dp[i] = Integer.MAX_VALUE;
for (int j = i, sum = 0; j <= n-k; ++j) {
sum += nums[j];
if (sum >= dp[i]) break;
dp[i] = Math.max(sum, dp[j+1]);
}
if (k == m) break;
}
}
return dp[0];
}``````
X. brute force
https://discuss.leetcode.com/topic/64189/java-recursive-dp-having-trouble-in-iterative-dp
``````public int splitArray(int[] nums, int m) {
if (nums.length == 0 || nums == null || m == 0)
return Integer.MAX_VALUE;
return splitArray(nums, m, 0);
}

public int splitArray(int[] nums, int m, int start) {
if (nums.length == 0 || nums == null || m == 0)
return Integer.MAX_VALUE;
if (start > nums.length)
return Integer.MAX_VALUE;
if (m == 1) {
int sum = 0;
for (int i = start; i < nums.length; i++)
sum += nums[i];
return sum;
}
int sum = 0;
int split = 0;
int min = Integer.MAX_VALUE;
for (int i = start; i < nums.length; i++) {
sum += nums[i];
split = Math.max(sum, splitArray(nums, m - 1, i + 1));
min = Math.min(min, split);
}
return min;
}``````

http://www.geeksforgeeks.org/split-array-two-equal-sum-subarrays/
Given an array of integers greater than zero, find if it is possible to split it in two subarrays (without reordering the elements), such that the sum of the two subarrays is the same.

An Efficient solution is to first compute the sum of the whole array from left to right. Now we traverse array from right and keep track of right sum, left sum can be computed by subtracting current element from whole sum.
`int` `findSplitPoint(``int` `arr[], ``int` `n)`
`{`
`    ``// traverse array element and compute sum`
`    ``// of whole array`
`    ``int` `leftSum = 0;`
`    ``for` `(``int` `i = 0 ; i < n ; i++)`
`        ``leftSum += arr[i];`
`    ``// again traverse array and compute right sum`
`    ``// and also check left_sum equal to right`
`    ``// sum or not`
`    ``int` `rightSum = 0;`
`    ``for` `(``int` `i=n-1; i >= 0; i--)`
`    ``{`
`        ``// add current element to right_sum`
`        ``rightSum += arr[i];`
`        ``// exclude current element to the left_sum`
`        ``leftSum -=  arr[i] ;`
`        ``if` `(rightSum == leftSum)`
`            ``return` `i ;`
`    ``}`
`    ``// if it is not possible to split array`
`    ``// into two parts.`
`    ``return` `-1;`
`}`

Simple solution is to run two loop to split array and check it is possible to split array into two parts such that sum of first_part equal to sum of second_part.
`int` `findSplitPoint(``int` `arr[], ``int` `n)`
`{`
`    ``int` `leftSum = 0 ;`
`    ``// traverse array element`
`    ``for` `(``int` `i = 0; i < n; i++)`
`    ``{`
`        ``// add current element to left Sum`
`        ``leftSum += arr[i] ;`
`        ``// find sum of rest array elements (rightSum)`
`        ``int` `rightSum = 0 ;`
`        ``for` `(``int` `j = i+1 ; j < n ; j++ )`
`            ``rightSum += arr[j] ;`
`        ``// split point index`
`        ``if` `(leftSum == rightSum)`
`            ``return` `i+1 ;`
`    ``}`
`    ``// if it is not possible to split array into`
`    ``// two parts`
`    ``return` `-1;`
`}`

http://www.geeksforgeeks.org/allocate-minimum-number-pages/
Given number of pages in n different books and m students. The books are arranged in ascending order of number of pages. Every student is assigned to read some consecutive books. The task is to assign books in such a way that the maximum number of pages assigned to a student is minimum.
The idea is to use Binary Search. We fix a value for number of pages as mid of current minimum and maximum. We initialize minimum and maximum as 0 and sum-of-all-pages respectively. If a current mid can be a solution, then we search on the lower half, else we search in higher half.
Now the question arises, how to check if a mid value is feasible or not? Basically we need to check if we can assign pages to all students in a way that the maximum number doesn’t exceed current value. To do this, we sequentially assign pages to every student while current number of assigned pages doesn’t exceed the value. In this process, if number of students become more than m, then solution is not feasible. Else feasible.
`// Utility function to check if current minimum value`
`// is feasible or not.`
`bool` `isPossible(``int` `arr[], ``int` `n, ``int` `m, ``int` `curr_min)`
`{`
`    ``int` `studentsRequired = 1;`
`    ``int` `curr_sum = 0;`
`    ``// iterate over all books`
`    ``for` `(``int` `i = 0; i < n; i++)`
`    ``{`
`        ``// check if current number of pages are greater`
`        ``// than curr_min that means we will get the result`
`        ``// after mid no. of pages`
`        ``if` `(arr[i] > curr_min)`
`            ``return` `false``;`
`        ``// count how many students are required`
`        ``// to distribute curr_min pages`
`        ``if` `(curr_sum + arr[i] > curr_min)`
`        ``{`
`            ``// increment student count`
`            ``studentsRequired++;`
`            ``// update curr_sum`
`            ``curr_sum = arr[i];`
`            ``// if students required becomes greater`
`            ``// than given no. of students,return false`
`            ``if` `(studentsRequired > m)`
`                ``return` `false``;`
`        ``}`
`        ``// else update curr_sum`
`        ``else`
`            ``curr_sum += arr[i];`
`    ``}`
`    ``return` `true``;`
`}`
`// function to find minimum pages`
`int` `findPages(``int` `arr[], ``int` `n, ``int` `m)`
`{`
`    ``long` `long` `sum = 0;`
`    ``// return -1 if no. of books is less than`
`    ``// no. of students`
`    ``if` `(n < m)`
`        ``return` `-1;`
`    ``// Count total number of pages`
`    ``for` `(``int` `i = 0; i < n; i++)`
`        ``sum += arr[i];`
`    ``// initialize start as 0 pages and end as`
`    ``// total pages`
`    ``int` `start = 0, end = sum;`
`    ``int` `result = INT_MAX;`
`    ``// traverse until start <= end`
`    ``while` `(start <= end)`
`    ``{`
`        ``// check if it is possible to distribute`
`        ``// books by using mid is current minimum`
`        ``int` `mid = (start + end) / 2;`
`        ``if` `(isPossible(arr, n, m, mid))`
`        ``{`
`            ``// if yes then find the minimum distribution`
`            ``result = min(result, mid);`
`            ``// as we are finding minimum and books`
`            ``// are sorted so reduce end = mid -1`
`            ``// that means`
`            ``end = mid - 1;`
`        ``}`
`        ``else`
`            ``// if not possible means pages should be`
`            ``// increased so update start = mid + 1`
`            ``start = mid + 1;`
`    ``}`
`    ``// at-last return minimum no. of  pages`
`    ``return` `result;`
`}`