**Range Query on a List - Google Top Interview Questions**

### Problem Statement :

Implement a data structure with the following methods: RangeSum(int[] nums) constructs a new instance with the given nums. total(int i, int j) returns the sum of integers from nums between [i, j). That is, nums[i] + nums[i + 1] + ... + nums[j - 1]. Constraints n ≤ 100,000 where n is the length of nums k ≤ 100,000 where k is the number of calls to total Example 1 Input methods = ["constructor", "total", "total"] arguments = [[[1, 2, 5, 0, 3, 7]], [0, 3], [1, 5]]` Output [None, 8, 10] Explanation r = RangeSum([1,2,5,0,3,7]) r.total(0, 3) == 8 # sum([1, 2, 5]) r.total(1, 5) == 10 # sum([2, 5, 0, 3])

### Solution :

` ````
Solution in C++ :
class RangeSum {
public:
RangeSum(vector<int>& nums) {
n = nums.size();
sum.resize(n, 0);
sum[0] = nums[0];
for (int i = 1; i < n; ++i) {
sum[i] = sum[i - 1] + nums[i];
}
}
int total(int i, int j) {
int range = INT_MIN;
if (i >= 0 && j >= 1 && i < n && j <= n && i <= j) {
int left = i > 0 ? sum[i - 1] : 0;
int right = sum[j - 1];
range = right - left;
}
return range;
}
private:
vector<int> sum;
int n;
};
```

` ````
Solution in Java :
import java.util.*;
class RangeSum {
int[] pf;
public RangeSum(int[] nums) {
pf = new int[nums.length];
pf[0] = nums[0];
for (int i = 1; i < nums.length; i++) {
pf[i] = nums[i] + pf[i - 1];
}
}
public int total(int i, int j) {
return i == 0 ? pf[j - 1] : pf[j - 1] - pf[i - 1];
}
}
```

` ````
Solution in Python :
class RangeSum:
def calculateprefixsum(self, vals):
vals.insert(0, 0)
for i in range(1, len(vals)):
vals[i] += vals[i - 1]
return vals
def __init__(self, nums):
self.nums = nums
self.prefixsum = self.calculateprefixsum(nums)
def total(self, i, j):
return self.prefixsum[j] - self.prefixsum[i]
```

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