Root Sum Sort — Problem Statement & Solution Guide
Problem Description
Sort an array of integers in ascending order based on the digital root, which is the value obtained by recursive sum of its digits until only a single digit remains.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Root Sum Sort"
WHY DOES IT MATTER?
Sorting by a derived key that has a small, fixed range is a classic case for counting or bucket sort, which offers linear time complexity and avoids the log factor of comparison sorts. This pattern is essential when the key domain is limited, as it guarantees optimal performance regardless of input size.
OPTIMIZATION CHALLENGE
The key insight is that the digital root’s domain is constant (10 values). By precomputing the root once per element and using a fixed‑size array of buckets, we eliminate repeated calculations and comparisons, reducing both time and space overhead.
REAL-WORLD CONNECTION
In distributed log aggregation, logs are often bucketed by severity level (e.g., INFO, WARN, ERROR). Since the number of levels is small, a bucket‑like approach allows efficient grouping and ordering without expensive comparisons, similar to sorting by digital root.
When explaining this in an interview, emphasize the constant‑size bucket array and the linear pass, and be ready to discuss how you would handle negative numbers or large integers to avoid overflow.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
Digital root is the single digit obtained by repeatedly summing the digits of a number until only one digit remains. For any integer n, its digital root can be computed in O(1) time using the congruence property: dr(n) = 1 + ((n-1) mod 9) for n>0, and 0 for n=0. A naive approach would compute the digital root for each element during a comparison in a standard comparison‑based sort, leading to O(n log n) time with repeated expensive digit summations. The optimal paradigm leverages the fact that digital roots are bounded between 0 and 9; we can bucket elements by their root and then concatenate the buckets. This reduces the sorting to a single linear pass for bucket assignment and a linear concatenation, achieving O(n) time and O(n) auxiliary space, while preserving stability if required.
Interview Questions on This Problem
Q1How would you sort an array by digital root in O(n) time, and why is this approach preferable over a standard sort?
Because digital roots are limited to 10 possible values (0–9), we can use counting or bucket sort. We iterate once to compute each element’s root (O(1) per element) and place it into the corresponding bucket. Finally, we concatenate the buckets in order. This is O(n) time versus O(n log n) for comparison sorts, and it avoids repeated root calculations during comparisons.
Q2What edge cases must you consider when implementing digital root sorting, especially for negative numbers or zeros?
Digital root is defined for non‑negative integers; for negative numbers we typically take the absolute value before computing the root, or define a convention. Zero has a digital root of 0, so it must be placed in the 0 bucket. Also, very large integers may overflow if not handled with 64‑bit types or big integers, so use appropriate data types.
Q3Explain how you would adapt this algorithm to a distributed system where the array is partitioned across multiple nodes. What challenges arise?
Each node can locally bucket its partition in O(partitionSize) time. After local bucketing, nodes must exchange bucket counts or merge buckets. The challenge is to preserve global order: we need to aggregate counts to determine the starting index of each bucket in the final array, then perform a parallel prefix sum to compute offsets, and finally scatter the elements to their correct positions. Communication overhead and load balancing are key concerns.
Examples
Input
[12, 13, 25, 67, 89]
Output
[12, 13, 25, 67, 89]
Explanation: Step-by-step: First, calculate the digital root for each number in the array: [12 -> 3, 13 -> 4, 25 -> 7, 67 -> 4, 89 -> 16]. Then, sort the numbers based on their digital roots: [3, 4, 4, 7, 16]. Finally, sort the numbers with the same digital root: [12, 13] and [25, 67].
Input
[9, 38, 100, 200]
Output
[9, 100, 38, 200]
Explanation: Step-by-step: First, calculate the digital root for each number in the array: [9 -> 9, 38 -> 2, 100 -> 1, 200 -> 2]. Then, sort the numbers based on their digital roots: [1, 2, 2, 9]. Finally, sort the numbers with the same digital root: [100] and [38, 200].
Constraints
- 1 <= array length <= 10^5
- 0 <= array elements <= 10^6
Optimal Approach & Strategy
Use a bucket sort: compute each number’s root once, place it into one of ten buckets, then concatenate the buckets. This runs in linear time and uses linear auxiliary space.
Brute Force Approach
Compute the digital root for each pair of numbers during a comparison sort, leading to O(n log n) time with repeated digit summations. This is inefficient for large arrays because each comparison does extra work.
Verified Code Solutions
function rootSumSort(nums) {
function digitalRoot(n) {
if (n < 10) return n;
return digitalRoot(n.toString().split('').reduce((a, b) => parseInt(a) + parseInt(b), 0));
}
return nums.sort((a, b) => {
let rootA = digitalRoot(a);
let rootB = digitalRoot(b);
if (rootA === rootB) {
return a - b;
}
return rootA - rootB;
});
}class Solution {
public:
vector<int> rootSumSort(vector<int>& nums) {
vector<int> result = nums;
sort(result.begin(), result.end(), [](int a, int b) {
int rootA = digitalRoot(a);
int rootB = digitalRoot(b);
if (rootA == rootB) {
return a - b;
}
return rootA - rootB;
});
return result;
}
int digitalRoot(int n) {
if (n < 10) return n;
return digitalRoot(sumOfDigits(n));
}
int sumOfDigits(int n) {
int sum = 0;
while (n > 0) {
sum += n % 10;
n /= 10;
}
return sum;
}
};class Solution {
public int[] rootSumSort(int[] nums) {
int[] result = nums.clone();
Arrays.sort(result, (a, b) -> {
int rootA = digitalRoot(a);
int rootB = digitalRoot(b);
if (rootA == rootB) {
return a - b;
}
return rootA - rootB;
});
return result;
}
private int digitalRoot(int n) {
if (n < 10) return n;
return digitalRoot(sumOfDigits(n));
}
private int sumOfDigits(int n) {
int sum = 0;
while (n > 0) {
sum += n % 10;
n /= 10;
}
return sum;
}
}def root_sum_sort(nums):
def digital_root(n):
if n < 10:
return n
return digital_root(sum(int(digit) for digit in str(n)))
return sorted(nums, key=lambda x: digital_root(x))
function rootSumSort(nums) {
function digitalRoot(n) {
if (n < 10) return n;
return digitalRoot(n.toString().split('').reduce((a, b) => parseInt(a) + parseInt(b), 0));
}
return nums.sort((a, b) => {
let rootA = digitalRoot(a);
let rootB = digitalRoot(b);
if (rootA === rootB) {
return a - b;
}
return rootA - rootB;
});
}Solve in Interative Editor
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