2578. Split With Minimum SumEasy Topics Companies Hint Given a positive integer num, split it into two non-negative integers num1 and num2 such that: The concatenation of num1 and num2 is a permutation of num. In other words, the sum of the number of occurrences of each digit in ...
where the largest sum among the two subarrays is only 18. 题解: Let dp[i][j] denotes up to index j, with i cut, the minimum of largest sum of subarrays. Thus with i = 0, that means there is no cut. Initialize dp[0][j] as sum of nums[0] to nums[j]. For dp[i][j],...
Since we need to compute subarray sum, so we should get the prefix sum array ps. Notice that nums[i] is nonnegative, so ps is non-decreasing. Using ps we can either do a linear scan with binary search, or just a linear scan. Solution 1. O(N*logN), fix the start index of the ...
classSolution{publicintsplitArray(int[]nums,intm){intn=nums.length;int[][]dp=newint[n+1][m+1];for(int[]row:dp){Arrays.fill(row,Integer.MAX_VALUE);// important}dp[0][0]=0;for(inti=1;i<=n;++i){for(intj=1;j<=Math.min(i,m);++j){intsum=0;for(intk=i;k>=j;--k){...