Sum of Sublist Range Sum - Google Top Interview Questions
Problem Statement :
You are given a list of non-negative integers nums and integers i and j. Let A be a sorted list of the sum of every sublist of nums. Return the sum of A from [i, j], inclusive. Constraints 1 ≤ n ≤ 100,000 where n is the length of nums Example 1 Input nums = [1, 2, 3, 4] i = 2 j = 3 Output 6 Explanation A = [1, 2, 3, 3, 4, 5, 6, 7, 9, 10] here and sum([3, 3]) = 6.
Solution :
Solution in C++ :
typedef long long ll;
int solve(vector<int>& nums, int i, int j) {
int n = nums.size();
vector<ll> pre(n + 1);
for (int i = 1; i <= n; ++i) pre[i] = pre[i - 1] + nums[i - 1];
vector<ll> prepre(n + 1);
for (int i = 1; i <= n; ++i) prepre[i] = prepre[i - 1] + pre[i];
auto count = [&](ll sum) {
ll ans = 0;
for (int i = 1; i <= n; ++i)
ans += upper_bound(pre.begin() + i, pre.end(), sum + pre[i - 1]) - (pre.begin() + i);
return ans;
};
auto calc = [&](ll p) {
ll l = 0, r = pre[n];
while (l <= r) {
ll mid = (l + r) >> 1;
if (count(mid) < p) {
l = mid + 1;
} else {
r = mid - 1;
}
}
ll smaller = count(l - 1);
ll equal = p - smaller;
ll sum = 0;
for (int i = 1; i <= n; ++i) {
int pos = upper_bound(pre.begin() + i, pre.end(), pre[i - 1] + l - 1) - pre.begin();
if (pos > i) sum += prepre[pos - 1] - prepre[i - 1] - pre[i - 1] * (pos - i);
}
return sum + equal * l;
};
return calc(j + 1) - calc(i);
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[] nums, int ii, int jj) {
int n = nums.length, res = 0, s = 0;
for (int a : nums) s += a;
long[] sum = new long[n], sum2 = new long[n];
sum[0] = sum2[0] = nums[0];
for (int i = 1; i < n; ++i) {
sum[i] = sum[i - 1] + nums[i];
sum2[i] = sum2[i - 1] + sum[i];
}
int low = findBound(nums, ii, s), high = findBound(nums, jj, s);
int countHigh = count(high, nums), countLow = count(low, nums);
res = low * Math.min(countLow - ii, jj - ii + 1);
++low;
if (low > high)
return res;
for (int a = -1, b = -1, k = 0, sa = 0, sb = 0; k < n; ++k) {
long pre = k > 0 ? sum[k - 1] : 0;
while (b + 1 < n && sum[b + 1] - pre <= high) ++b;
while (a < k || (a <= b && sum[a] - pre < low)) ++a;
if (a <= b) {
long t = sum2[b] - (a > 0 ? sum2[a - 1] : 0);
if (k > 0)
t -= sum[k - 1] * (b - a + 1);
res += (int) t;
}
}
if (countHigh > jj + 1)
res -= high * (countHigh - jj - 1);
return res;
}
private int findBound(int[] ns, int idx, int max) {
int l = 0, r = max, mid;
while (l <= r) {
mid = (l + r) >> 1;
int t = count(mid, ns);
if (t >= idx + 1)
r = mid - 1;
else
l = mid + 1;
}
return l;
}
private int count(int v, int[] ns) {
int s = 0, res = 0, p = 0;
for (int i = 0; i < ns.length; ++i) {
s += ns[i];
while (s > v && p <= i) {
s -= ns[p++];
}
res += i - p + 1;
}
return res;
}
}
Solution in Python :
class Solution:
def solve(self, nums, i, j):
n = len(nums)
a = list(itertools.accumulate([0] + nums))
b = list(itertools.accumulate([0] + a))
def cnt(ub):
c, u = 0, 0
for v in range(1, n + 1):
while a[v] - a[u] > ub:
u += 1
c += v - u
return c
def go(m):
lo, hi = 0, a[n]
while lo < hi:
mid = (lo + hi) // 2
if cnt(mid) < m:
lo = mid + 1
else:
hi = mid
s, u = 0, 0
for v in range(1, n + 1):
while a[v] - a[u] > lo:
u += 1
s += a[v] * (v - u) - (b[v] - b[u])
return s - (cnt(lo) - m) * lo
return go(j + 1) - go(i)
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