Count of Range Sum in C++

Suppose we have an integer array nums, we have to find the number of range sums that lie in range [lower, upper] both inclusive. The range sum S(i, j) is defined as the sum of the elements in nums from index i to index j where i ≤ j.

So if the input is like [-3,6,-1], lower = -2 and upper = 2, then the result will be 2, as the ranges are [0,2], the sum is 2, [2,2], sum is -2.

To solve this, we will follow these steps −

• Define a function mergeIt(), this will take array prefix, start, mid, end, lower, upper,
• i := start, j := mid + 1
• temp := end - start + 1
• low := mid + 1, high := mid + 1
• k := 0
• Define an array arr of size: temp.
• while i <= mid, do −
  • while (low <= end and prefix[low] - prefix[i] < lower), do −
    • increase low by 1
  • while (high <= end and prefix[high] - prefix[i] <= upper), do −
    • increase high by 1
  • while (j <= end and prefix[j] < prefix[i]), do −
    • arr[k] := prefix[j]
    • (increase j by 1)
    • (increase k by 1)
  • arr[k] := prefix[i]
  • (increase i by 1)
  • (increase k by 1)
  • count := count + high - low
• while j <= end, do −
  • arr[k] := prefix[j]
  • (increase k by 1)
  • (increase j by 1)
• for initialize i := 0, when i < temp, update (increase i by 1), do −
  • prefix[start] := arr[i]
  • (increase start by 1)
• Define a function merge(), this will take prefix[], start, end, lower, upper,
  • if start >= end, then return
• mid := start + (end - start)
• call the function merge(prefix, start, mid, lower, upper)
• call the function merge(prefix, mid + 1, end, lower, upper)
• call the function mergeIt(prefix, start, mid, end, lower, upper)
• From the main method, do the following −
• n := size of nums
• count := 0
• Define an array prefix of size: n+1.
• prefix[0] := 0
• for initialize i := 1, when i <= n, update (increase i by 1), do −
  • prefix[i] := prefix[i - 1] + nums[i - 1]
• call the function merge(prefix, 0, n, lower, upper)
• return count

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h> using namespace std; typedef long long int lli; class Solution { public:    int count = 0;    void mergeIt(lli prefix[], lli start ,lli mid, lli end, lli lower, lli upper){       lli i = start, j = mid + 1;       lli temp = end - start + 1;       lli low = mid + 1, high = mid + 1;       lli k = 0;       lli arr[temp];       while(i <= mid){          while(low <= end && prefix[low] - prefix[i] < lower) low++;          while(high <= end && prefix[high] - prefix[i] <= upper) high++;          while(j<= end && prefix[j] < prefix[i]){             arr[k] = prefix[j];             j++;             k++;          }          arr[k] = prefix[i];          i++;          k++;          count += high - low;       }       while(j <= end){          arr[k] = prefix[j];          k++;          j++;       }       for(i = 0; i < temp; i++){          prefix[start] = arr[i];          start++;       }    }    void merge(lli prefix[], lli start, lli end, lli lower, lli upper){       if(start >= end)return;       lli mid = start + (end - start) / 2;       merge(prefix, start, mid, lower, upper);       merge(prefix, mid + 1, end, lower, upper);       mergeIt(prefix, start, mid, end, lower, upper);    }    int countRangeSum(vector<int>& nums, int lower, int upper) {       lli n = nums.size();       count = 0;       lli prefix[n + 1];       prefix[0] = 0;       for(lli i = 1; i <= n; i++){          prefix[i] = prefix[i - 1] + nums[i - 1];       }       merge(prefix, 0, n, lower, upper);       return count;    } }; main(){    Solution ob;    vector<int> v = {-3,6,-1};    cout << (ob.countRangeSum(v, -2, 2)); }

Input

{-3,6,-1} -2 2

Output

2