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# Check if Number Has Equal Digit Count and Digit Value LeetCode Solution

## Problem – Check if Number Has Equal Digit Count and Digit Value LeetCode Solution

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 = '1'. The digit 0 occurs once in num.
num = '2'. The digit 1 occurs twice in num.
num = '1'. The digit 2 occurs once in num.
num = '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'. The digit 0 should occur zero times, but actually occurs twice in num.
num = '3'. The digit 1 should occur three times, but actually occurs zero times in num.
num = '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.

### Check if Number Has Equal Digit Count and Digit Value LeetCode Solution in Java

``````class Solution {
public boolean digitCount(String num) {
int[] freqArr = new int;  // n = 10 given in constraints;

for(char ch : num.toCharArray()){
freqArr[ch-'0']++;
}

for(int i=0;i<num.length();i++){
int freq = num.charAt(i)-'0';  //freq of each indexValue;
freqArr[i] = freqArr[i] - freq;
}
for(int i=0;i<10;i++){
if(freqArr[i]!=0){
return false;
}
}
return true;
}
}``````

### Check if Number Has Equal Digit Count and Digit Value LeetCode Solution in C++

``````class Solution {
public:
bool digitCount(string num) {
unordered_map<char, int> mpp;

for(int i=0; i<num.size(); i++)
mpp[num[i]]++;

for(int i=0; i<num.size(); i++){
char c ='0' + i;
// cout<<c<<endl;
if(num[i] != '0' + mpp[c]){
// cout<<num[i]<<" "<<mpp[c]<<endl;
return false;
}

}

return true;
}
};
``````

### Check if Number Has Equal Digit Count and Digit Value LeetCode Solution in Python

``````class Solution:
def digitCount(self, num: str) -> bool:
n=len(num)

for i in range(n):
if(num.count(str(i))!=int(num[i])):
return False

return True``````
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