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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 (low <= end and prefix[low] - prefix[i] < lower), do −
- 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
#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
2Output
2
Updated on: 01-Jun-2020
425 Views
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