2283-CheckIfNumberHasEqualDigitCountAndDigitValue.go

package main

// 2283. Check if Number Has Equal Digit Count and Digit Value
// You are given a 0-indexed string num of length n consisting of digits.

// Return true if for every index i in the range 0 <= i < n, the digit i occurs num[i] times in num, otherwise return false.

// Example 1:
// Input: num = "1210"
// Output: true
// Explanation:
// num[0] = '1'. The digit 0 occurs once in num.
// num[1] = '2'. The digit 1 occurs twice in num.
// num[2] = '1'. The digit 2 occurs once in num.
// num[3] = '0'. The digit 3 occurs zero times in num.
// The condition holds true for every index in "1210", so return true.

// Example 2:
// Input: num = "030"
// Output: false
// Explanation:
// num[0] = '0'. The digit 0 should occur zero times, but actually occurs twice in num.
// num[1] = '3'. The digit 1 should occur three times, but actually occurs zero times in num.
// num[2] = '0'. The digit 2 occurs zero times in num.
// The indices 0 and 1 both violate the condition, so return false.

// Constraints:
//     n == num.length
//     1 <= n <= 10
//     num consists of digits.

import "fmt"

func digitCount(num string) bool {
    mp := make([]int, 10)
    for _, v := range num {
        mp[v - '0']++
    }
    for i, v := range num {
        if mp[i] != int(v - '0') {
            return false
        }
    }
    return true
}

func main() {
    // Example 1:
    // Input: num = "1210"
    // Output: true
    // Explanation:
    // num[0] = '1'. The digit 0 occurs once in num.
    // num[1] = '2'. The digit 1 occurs twice in num.
    // num[2] = '1'. The digit 2 occurs once in num.
    // num[3] = '0'. The digit 3 occurs zero times in num.
    // The condition holds true for every index in "1210", so return true.
    fmt.Println(digitCount("1210")) // true
    // Example 2:
    // Input: num = "030"
    // Output: false
    // Explanation:
    // num[0] = '0'. The digit 0 should occur zero times, but actually occurs twice in num.
    // num[1] = '3'. The digit 1 should occur three times, but actually occurs zero times in num.
    // num[2] = '0'. The digit 2 occurs zero times in num.
    // The indices 0 and 1 both violate the condition, so return false.
    fmt.Println(digitCount("030")) // false

    fmt.Println(digitCount("111111111")) // false
    fmt.Println(digitCount("123456789")) // false
}