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3723. Maximize Sum of Squares of Digits

Description

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

Solutions

Solution 1: Greedy

If $\text{num} \times 9 < \text{sum}$, then there is no valid good integer, so we return an empty string.

Otherwise, we can use as many digits $9$ as possible to form the good integer, since $9^2$ is the largest and will maximize the score. Specifically, we calculate how many $9$s are contained in $\text{sum}$, denoted as $k$, and the remaining part $s = \text{sum} - 9 \times k$. Then, we construct the good integer with the first $k$ digits as $9$. If $s > 0$, we append a digit $s$ at the end, and finally pad with $0$s to reach a total of $\text{num}$ digits.

The time complexity is $O(\text{num})$ and the space complexity is $O(\text{num})$, where $\text{num}$ is the number of digits in the good integer.

• Java
• C++
• Python
• Go
• TypeScript

• 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();
      }
  }
  
  

• class Solution {
  public:
      string maxSumOfSquares(int num, int sum) {
          if (num * 9 < sum) {
              return "";
          }
          int k = sum / 9, s = sum % 9;
          string ans(k, '9');
          if (s > 0) {
              ans += char('0' + s);
          }
          ans += string(num - ans.size(), '0');
          return ans;
      }
  };
  
  

• 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
  
  

• 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
  }
  
  

• 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;
  }
  
  

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