Sum of Integers with Maximum Digit Range solution and explanation
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The problem
You are given an integer array nums.
The digit range of an integer is defined as the difference between its largest digit and smallest digit.
For example, the digit range of 5724 is 7 - 2 = 5.
Return the sum of all integers in nums whose digit range is equal to the maximum digit range among all integers in the array.
Example 1:
Input: nums = [5724,111,350]
Output: 6074
Explanation:
i |
nums[i] |
Largest | Smallest | Digit Range |
|---|---|---|---|---|
| 0 | 5724 | 7 | 2 | 5 |
| 1 | 111 | 1 | 1 | 0 |
| 2 | 350 | 5 | 0 | 5 |
The maximum digit range is 5. The integers with this digit range are 5724 and 350, so the answer is 5724 + 350 = 6074.
Example 2:
Input: nums = [90,900]
Output: 990
Explanation:
i |
nums[i] |
Largest | Smallest | Digit Range |
|---|---|---|---|---|
| 0 | 90 | 9 | 0 | 9 |
| 1 | 900 | 9 | 0 | 9 |
The maximum digit range is 9. Both integers have this digit range, so the answer is 90 + 900 = 990.
Constraints:
1 <= nums.length <= 10010 <= nums[i] <= 105
Approach overview
Problem Overview: You need to identify integers whose digit range is the largest among all numbers in the input. The digit range is typically computed as maxDigit - minDigit for each integer. After finding the maximum range, sum every integer that matches it.
Approach 1: Brute Force Digit Comparison (O(n * k), O(1))
Iterate through every number and extract its digits one by one using modulo and division operations. For each integer, track the smallest and largest digit encountered, then compute the digit range. Store the largest range seen so far and update the running sum accordingly. This approach works well because digit extraction is cheap, and each number contains at most k digits. Problems involving repeated digit processing often appear under Math and Implementation categories.
Approach 2: Single Pass Optimized Tracking (O(n * k), O(1))
You can optimize the logic by combining range calculation and result aggregation in a single traversal. As you process each integer, compute its digit range immediately. If the range exceeds the current maximum, reset the accumulated sum to the current number. If the range matches the maximum, add the number to the sum. This avoids storing intermediate arrays or maps and keeps memory usage constant. The key insight is that you only care about the current best digit range, not every previously computed value.
Approach 3: Precomputed Digit Statistics (O(n * k), O(n))
For repeated queries on the same dataset, you can precompute the digit range for every number and cache the results in an auxiliary array. This makes later aggregations or filtering operations faster because the expensive digit extraction step runs once. The tradeoff is additional memory usage proportional to the number of integers. This pattern is common in Array preprocessing problems where multiple passes are expected.
Recommended for interviews: Interviewers usually expect the single-pass tracking approach because it demonstrates efficient state management and clean implementation. Starting with the brute force explanation shows you understand the underlying digit operations, while the optimized pass proves you can reduce unnecessary storage and combine computations effectively.
Solution
We traverse the array nums. For each integer x, we extract its digits to find the largest digit b and the smallest digit a, then compute the digit range r = b - a. If r is greater than the current maximum digit range mx, we update mx = r and reset the answer to x; if r equals mx, we add x to the answer.
The time complexity is O(n log M), and the space complexity is O(1), where n is the length of the array nums and M is the maximum value in the array.
class Solution:
def maxDigitRange(self, nums: list[int]) -> int:
ans = mx = 0
for x in nums:
a, b = 10, 0
y = x
while y:
v = y % 10
y //= 10
a = min(a, v)
b = max(b, v)
r = b - a
if mx < r:
mx = r
ans = x
elif mx == r:
ans += x
return ansclass Solution {
public int maxDigitRange(int[] nums) {
int ans = 0, mx = 0;
for (int x : nums) {
int a = 10, b = 0;
for (int y = x; y > 0; y /= 10) {
int v = y % 10;
a = Math.min(a, v);
b = Math.max(b, v);
}
int r = b - a;
if (mx < r) {
mx = r;
ans = x;
} else if (mx == r) {
ans += x;
}
}
return ans;
}
}class Solution {
public:
int maxDigitRange(vector<int>& nums) {
int ans = 0, mx = 0;
for (int x : nums) {
int a = 10, b = 0;
for (int y = x; y > 0; y /= 10) {
int v = y % 10;
a = min(a, v);
b = max(b, v);
}
int r = b - a;
if (mx < r) {
mx = r;
ans = x;
} else if (mx == r) {
ans += x;
}
}
return ans;
}
};function maxDigitRange(nums: number[]): number {
let [ans, mx] = [0, 0];
for (const x of nums) {
let [a, b] = [10, 0];
for (let y = x; y; y = (y / 10) | 0) {
const v = y % 10;
a = Math.min(a, v);
b = Math.max(b, v);
}
const r = b - a;
if (mx < r) {
mx = r;
ans = x;
} else if (mx == r) {
ans += x;
}
}
return ans;
}func maxDigitRange(nums []int) (ans int) {
mx := 0
for _, x := range nums {
a, b := 10, 0
for y := x; y > 0; y /= 10 {
v := y % 10
a = min(a, v)
b = max(b, v)
}
r := b - a
if mx < r {
mx = r
ans = x
} else if mx == r {
ans += x
}
}
return
}Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force Digit Comparison | O(n * k) | O(1) | Best for straightforward implementations and interviews |
| Single Pass Optimized Tracking | O(n * k) | O(1) | General optimal solution with constant extra memory |
| Precomputed Digit Statistics | O(n * k) | O(n) | Useful when multiple queries reuse the same numbers |
Frequently asked questions
Is Sum of Integers with Maximum Digit Range easy or hard?
The problem is generally categorized as Easy because the core logic relies on simple digit traversal and comparison. The main challenge is correctly updating the maximum range and handling ties while maintaining the final sum.
Sum of Integers with Maximum Digit Range Python/Java solution
Python solutions usually use modulo and floor division to extract digits efficiently inside a loop. Java implementations follow the same logic with integer arithmetic and constant extra memory, making both languages suitable for an O(n * k) implementation.
How to solve Sum of Integers with Maximum Digit Range in O(n)?
If the maximum number of digits is treated as a constant, the solution behaves like O(n). Iterate through the array once, compute the digit range for every number, update the current maximum range, and maintain the running sum for matching integers.
What is the best approach for Sum of Integers with Maximum Digit Range?
The best approach is a single-pass traversal that computes the digit range for each number while tracking the current maximum range and accumulated sum. It runs in O(n * k) time, where k is the number of digits per integer, and uses O(1) extra space.
Is Sum of Integers with Maximum Digit Range asked at Google/Amazon/Meta?
Digit-processing and number-analysis problems appear frequently in coding interviews at companies like Amazon, Google, and Meta. Variants often test implementation accuracy, edge-case handling, and optimization of repeated numeric operations.
What data structure is used in Sum of Integers with Maximum Digit Range?
The problem typically does not require advanced data structures. Most solutions rely on integer arithmetic, loop traversal, and a few scalar variables for tracking the current maximum digit range and result sum.
What is the time complexity of Sum of Integers with Maximum Digit Range?
The optimal solution runs in O(n * k) time because each integer must be processed digit by digit. Space complexity is O(1) since only a few tracking variables are needed during traversal.