Counting the number of duplicates with nested loops. How to escape from comparing two chars with the same index

Question

Function that will return the count of distinct case-insensitive alphabetic characters and numeric digits that occur more than once in the input string.

"abcde" -> 0 # no characters repeats more than once

"indivisibility" -> 1 # 'i' occurs six times rs twice

The problem is that each iteration loops meet on the same char and compare it. How can I avoid it?

function duplicateCount(text){
  var texT = text.toLowerCase();
var count = 0;
  var total = 0;
  for(var i = 0; i < texT.length; i++ ){
      var char = texT[i];
      if(count > 1){
            total = total + 1;
      }
  
      for ( var j = 0; j < texT.length; j++){
          var char2 = texT[j];
          if(char === char2){
              count = count + 1;
          }
    
          
      }
      
  }
     return total;
  
}
duplicateCount('kBHhJkj8l8');

Answer 1

function duplicateCount(text){
  var texT = text.toLowerCase();
  const obj = {};
  for(var i = 0; i < texT.length; i++ ){
      if(obj[texT[i]]) {
        obj[texT[i]] += 1;
      }
      else {
        obj[texT[i]] = 1;
      }
  }
  let total = 0;
  Object.keys(obj).forEach(key => {
    if(obj[key] != 1){
      total += 1;
    }
  })
  return total;
}
console.log(duplicateCount('kBHhJkj8l8'));