LeetCode #3723 — MEDIUM

Maximize Sum of Squares of Digits

Move from brute-force thinking to an efficient approach using math strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given two positive integers num and sum.

A positive integer n is good if it satisfies both of the following:

The number of digits in n is exactly num.
The sum of digits in n is exactly sum.

The score of a good integer n is the sum of the squares of digits in n.

Return a string denoting the good integer n that achieves the maximum score. If there are multiple possible integers, return the maximum one. If no such integer exists, return an empty string.

Example 1:

Input: num = 2, sum = 3

Output: "30"

Explanation:

There are 3 good integers: 12, 21, and 30.

The score of 12 is 1² + 2² = 5.
The score of 21 is 2² + 1² = 5.
The score of 30 is 3² + 0² = 9.

The maximum score is 9, which is achieved by the good integer 30. Therefore, the answer is "30".

Example 2:

Input: num = 2, sum = 17

Output: "98"

Explanation:

There are 2 good integers: 89 and 98.

The score of 89 is 8² + 9² = 145.
The score of 98 is 9² + 8² = 145.

The maximum score is 145. The maximum good integer that achieves this score is 98. Therefore, the answer is "98".

Example 3:

Input: num = 1, sum = 10

Output: ""

Explanation:

There are no integers that have exactly 1 digit and whose digits sum to 10. Therefore, the answer is "".

Constraints:

1 <= num <= 2 * 10⁵
1 <= sum <= 2 * 10⁶

Step 01

Brute Force Baseline

Problem summary: You are given two positive integers num and sum. A positive integer n is good if it satisfies both of the following: The number of digits in n is exactly num. The sum of digits in n is exactly sum. The score of a good integer n is the sum of the squares of digits in n. Return a string denoting the good integer n that achieves the maximum score. If there are multiple possible integers, return the maximum one. If no such integer exists, return an empty string.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Greedy

Example 1

2
3

Example 2

2
17

Example 3

1
10

Step 02

Core Insight

What unlocks the optimal approach

Use a greedy approach.
Fill the leftmost digits with 9s first (as much as possible).

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3723: Maximize Sum of Squares of Digits
class Solution {
    public String maxSumOfSquares(int num, int sum) {
        if (num * 9 < sum) {
            return "";
        }
        int k = sum / 9;
        sum %= 9;
        StringBuilder ans = new StringBuilder("9".repeat(k));
        if (sum > 0) {
            ans.append(sum);
        }
        ans.append("0".repeat(num - ans.length()));
        return ans.toString();
    }
}

// Accepted solution for LeetCode #3723: Maximize Sum of Squares of Digits
func maxSumOfSquares(num int, sum int) string {
	if num*9 < sum {
		return ""
	}

	k, s := sum/9, sum%9
	ans := strings.Repeat("9", k)
	if s > 0 {
		ans += string('0' + byte(s))
	}
	if len(ans) < num {
		ans += strings.Repeat("0", num-len(ans))
	}

	return ans
}

# Accepted solution for LeetCode #3723: Maximize Sum of Squares of Digits
class Solution:
    def maxSumOfSquares(self, num: int, sum: int) -> str:
        if num * 9 < sum:
            return ""
        k, s = divmod(sum, 9)
        ans = "9" * k
        if s:
            ans += digits[s]
        ans += "0" * (num - len(ans))
        return ans

// Accepted solution for LeetCode #3723: Maximize Sum of Squares of Digits
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3723: Maximize Sum of Squares of Digits
// class Solution {
//     public String maxSumOfSquares(int num, int sum) {
//         if (num * 9 < sum) {
//             return "";
//         }
//         int k = sum / 9;
//         sum %= 9;
//         StringBuilder ans = new StringBuilder("9".repeat(k));
//         if (sum > 0) {
//             ans.append(sum);
//         }
//         ans.append("0".repeat(num - ans.length()));
//         return ans.toString();
//     }
// }

// Accepted solution for LeetCode #3723: Maximize Sum of Squares of Digits
function maxSumOfSquares(num: number, sum: number): string {
    if (num * 9 < sum) {
        return '';
    }

    const k = Math.floor(sum / 9);
    const s = sum % 9;

    let ans = '9'.repeat(k);
    if (s > 0) {
        ans += String.fromCharCode('0'.charCodeAt(0) + s);
    }
    if (ans.length < num) {
        ans += '0'.repeat(num - ans.length);
    }

    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(\textnum)

Space

O(\textnum)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.