Multiply Strings

Sun, 12 Apr 2026 00:00:00 +0000

43. Multiply Strings

Algorithm Steps

Iterate the array in reverse order
- Convert each character of num1 and num2 to integer, and store conversion results in integer array A and B
Simulate multiplication
- When multiplying 2 numbers of length n and m, the result has at most n + m digits
- Multiply A and B bit by bit accumulation，and store result in array C
  - C[i + j] = A[i] * B[j]
Handle carries
- Each digit of C is treated as unit digit, and the carry-over is accumulated to the next digit
Remove leading 0 and generate the result string
- Search for the first non-zero bit in the C from higher bit to the lower bit
- Convert the result array to string

Complexity Analysis

Length of num1 and num2 is n and m
Simulate multiplication O(n * m)
Handel carries and generate result string O(n + m)

Overall time complexity O(n * m)

Length of Result array C is n + m，space complexity O(n + m)

C++ Code

class Solution {
public:
    string multiply(string num1, string num2)
    {
        // get the length of num1 and num2
        int n = num1.size(), m = num2.size();

        // iterate the string in reverse order and convert to integer array
        std::vector<int> A, B;
        for (int i = n - 1; i >= 0; --i)
            A.push_back(num1[i] - '0');
        for (int i = m - 1; i >= 0; --i)
            B.push_back(num2[i] - '0');

        // initialize result array C
        std::vector<int> C(n + m, 0);

        // simulate multiplication
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < m; ++j)
                C[i + j] += A[i] * B[j];

        // handle carries
        for (int i = 0, t = 0; i < C.size(); ++i)
        {
            t += C[i];
            C[i] = t % 10;  // keep unit digit
            t /= 10;        // process carry part
        }

        // remove leading zeros
        int k = C.size() - 1;
        while (k > 0 && C[k] == 0) --k;

        // convert integer array to string
        std::string res;
        while (k >= 0)
            res += C[k--] + '0';

        return res;
    }
};

Python Code

class Solution:
    def multiply(self, num1: str, num2: str) -> str:
        n, m = len(num1), len(num2)

        A = [int(num1[i]) for i in range(n - 1, -1, -1)]
        B = [int(num2[i]) for i in range(m - 1, -1, -1)]

        C = [0] * (n + m)
        for i in range(n):
            for j in range(m):
                C[i + j] += A[i] * B[j]

        t = 0
        for i in range(len(C)):
            t += C[i]
            C[i] = t % 10
            t //= 10

        k = len(C) - 1
        while k > 0 and C[k] == 0:
            k -= 1

        res = ""
        while k >= 0:
            res += str(C[k])
            k -= 1

        return res

Go Code

func multiply(num1 string, num2 string) string {
    n, m := len(num1), len(num2)

    A := make([]int, n)
    B := make([]int, m)


    for i := 0; i < n; i ++ {
        A[i] = int(num1[n - 1 - i] - '0')
    }

    for i := 0; i < m; i ++ {
        B[i] = int(num2[m - 1 - i] - '0')
    }

    C := make([]int, n + m)
    for i := range A {
        for j := range B {
            C[i + j] += A[i] * B[j]
        }
    }

    t := 0
    for i := 0; i < len(C); i ++ {
        t += C[i]
        C[i] = t % 10
        t /= 10
    }

    k := len(C) - 1
    for k > 0 && C[k] == 0 {
        k -= 1
    }

    res := ""
    for k >= 0 {
        res += string(C[k] + '0')
        k -= 1
    }

    return res
}

JavaScript Code

/**
 * @param {string} num1
 * @param {string} num2
 * @return {string}
 */
var multiply = function (num1, num2) {
  let n = num1.length,
    m = num2.length;
  let A = [],
    B = [];

  for (let i = n - 1; i >= 0; i--) {
    A.push(Number(num1[i]));
  }

  for (let i = m - 1; i >= 0; i--) {
    B.push(Number(num2[i]));
  }

  let C = new Array(n + m).fill(0);
  for (let i = 0; i < n; i++) {
    for (let j = 0; j < m; j++) {
      C[i + j] += A[i] * B[j];
    }
  }

  let t = 0;
  for (let i = 0; i < C.length; i++) {
    t += C[i];
    C[i] = t % 10;
    t = Math.trunc(t / 10);
  }

  let k = C.length - 1;
  while (k > 0 && C[k] === 0) {
    k--;
  }

  let res = "";
  while (k >= 0) {
    res += String(C[k--]);
  }

  return res;
};

Minimum Distance Between Three Equal Elements I

Sat, 11 Apr 2026 00:00:00 +0000

3740. Minimum Distance Between Three Equal Elements I

Algorithm Steps

Use map<int, vector<int>> to record all the indices of each element
For each index array p
- Enumerate the length >= 3
- Calculate (p[i] - p[i-2]) * 2
Take the minimum value

Complexity Analysis

Time complexity: O(n)

Space complexity: O(n)

Cpp Code

class Solution {
public:
    int minimumDistance(vector<int>& nums) {
        std::unordered_map<int, std::vector<int>> f;
        for (int i = 0; i < nums.size(); ++ i)
            f[nums[i]].push_back(i); // record all the indices of each element

        int res = INT_MAX;
        for (auto [_, p] : f) // iterate each index array
            for (int i = 2; i < p.size(); ++ i)
                res = std::min(res, (p[i] - p[i - 2]) * 2); // calculate minimum distance = 2 * (k - i)

        return res == INT_MAX ? -1 : res;
    }
};

Python Code

class Solution:
    def minimumDistance(self, nums: List[int]) -> int:
        f = {}
        for i, x in enumerate(nums):
            if x not in f:
                f[x] = []
            f[x].append(i)

        res = float('inf')
        for p in f.values():
            for i in range(2, len(p)):
                res = min(res, (p[i] - p[i - 2]) * 2)

        return -1 if res == float('inf') else res

Go Code

func minimumDistance(nums []int) int {
    f := make(map[int][]int)
    for i := 0; i < len(nums); i ++ {
        f[nums[i]] = append(f[nums[i]], i)
    }

    INF := 0x3f3f3f3f
    res := INF
    for _, p := range f {
        for i := 2; i < len(p); i ++ {
            if (p[i] - p[i - 2]) * 2 < res {
                res = (p[i] - p[i - 2]) * 2
            }
        }
    }

    if res == INF {
        return -1
    }

    return res
}

JavaScript Code

/**
 * @param {number[]} nums
 * @return {number}
 */
var minimumDistance = function (nums) {
  let f = new Map();
  for (let i = 0; i < nums.length; i++) {
    if (!f.has(nums[i])) f.set(nums[i], []);
    f.get(nums[i]).push(i);
  }

  let res = Infinity;
  for (let p of f.values()) {
    for (let i = 2; i < p.length; i++) {
      res = Math.min(res, (p[i] - p[i - 2]) * 2);
    }
  }

  return res == Infinity ? -1 : res;
};

Go on Hang Zhang

Multiply Strings

43. Multiply Strings

Algorithm Steps

Complexity Analysis

C++ Code

Python Code

Go Code

JavaScript Code

Minimum Distance Between Three Equal Elements I

3740. Minimum Distance Between Three Equal Elements I

Algorithm Steps

Complexity Analysis

Cpp Code

Python Code

Go Code

JavaScript Code