leetcode@ [327] Count of Range Sum (Binary Search)

https://leetcode.com/problems/count-of-range-sum/

Given an integer array nums , return the number of range sums this lie in [lower, upper] inclusive.
Range sum S(i, j) nums i j is defined i as the sum of the elements in between indices and (≤ j ), inclusive.

Note:
A naive algorithm of O(n2) is trivial. You must does better than that.

Example:
Given nums = [-2, 5, -1] , lower = -2 , upper = 2 ,
Return 3 .
The three ranges is: [0, 0] , [2, 2] , and [0, 2] their respective sums are: -2, -1, 2 .

classPair { Public intidx;  Public LongVal;  PublicPairintIdxLongval) {        Super();  This. idx =idx;  This. val =Val; }    }classPaircomparatorImplementsComparator { Public intCompare (Object O1, Object O2) {pair P1=(pair) O1; Pair P2=(pair) O2; if(P1.val <p2.val) {return-1; } Else {            return1; }    }    } Public classSolution { Public StaticArraylist<pair> Tosortedlist (Long[] sum) {ArrayList<pair> ls =NewArraylist<pair> ();  for(inti=0; i<sum.length; ++i) {pair P=Newpair (I, sum[i]);        Ls.add (P); } collections.sort (LS,Newpaircomparator ()); returnls; }         Public Static intBinarySearch (arraylist<pair> ls,intLintRLongLbLongUbintindex) {                if(L >r) {return0; }        if(L = =r) {if(Ls.get (L). val >= lb && ls.get (l). Val <= ub && Ls.get (L). IDX >=index) {                //System.out.println ("Candidate Index range: [" + Index + "," + Ls.get (L). idx + "]");                return1; }            return0; }                intrs = 0; intMid = (L + r)/2; if(Ls.get (mid). val <lb) {RS= BinarySearch (ls, mid+1, R, LB, ub, index); } Else if(Ls.get (mid). val >UB) {RS= BinarySearch (LS, L, mid-1, LB, ub, index); } Else{RS= BinarySearch (LS, l, mid-1, LB, UB, index) + binarysearch (LS, mid+1, R, LB, ub, index); if(Ls.get (mid). IDX >=index) {                //System.out.println ("Candidate Index range: [" + Index + "," + Ls.get (L). idx + "]");rs++; }        }                returnrs; }         Public intCountrangesum (int[] Nums,intLowerintUpper) {                intn =nums.length; if(n = = 0) {            return0; }                Long[] sum =New Long[n]; sum[0] = Nums[0];  for(intI=1; i<n; ++i) {Sum[i]= Sum[i-1] +Nums[i]; }                intrs = 0; ArrayList<pair> ls =tosortedlist (sum);  for(inti=0; i<n; ++i) {LongNew_lower = (Long) Lower + sum[i]-(Long) nums[i]; LongNew_upper = (Long) Upper + sum[i]-(Long) nums[i]; intCount = BinarySearch (ls, 0, ls.size ()-1, New_lower, New_upper, i); RS+=count; }                returnrs; }}

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[email protected] [327] Count of Range Sum (Binary Search)