Range Query on a List - Google Top Interview Questions

                        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]