Count of Range Sum in C++

C++Server Side ProgrammingProgramming

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
raja
Published on 01-Jun-2020 10:36:30
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