Question 178584 in Jquery Question posted in
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Jquery – How can I replace element of array with X

Please I’m trying to find minimum value inside my array and replace the minimum value with X for the same element not to be picked as minimum over again

Here is what I did
Var arr = [3,8,2,5,1]; Var minIndex;

While(arr.indexOf(‘X’) === -1){
minIndex = 0;
$.each(arr,function(index,value){
if (value < arr.[minIndex]){
minIndex = index;
}
});
arr[minIndex] = ‘X’;

}
Console.log(arr);

But only the first minimum which is 1 and is replaced with X

I don’t know what I’m doing wrong
Any help please

Replacing element of array with X

Tags:

2

Answers


  1. 0

    You could use Array::reduce(). As mentioned by the author he wants to skip the minimum value for the next iteration, I guess we could remember the found minimum index:

    const arr = [3,8,2,5,1];

// find the minimum value's index
const [, minIdx] = arr.reduce(
  (acc, curr, idx) => acc[0] === undefined || curr < acc[0] ? [curr, idx] : acc,
  []
);

for(let i = 0; i < arr.length; i++){
  if(i === minIdx){
    continue; // skip minimum value
  }
  const val = arr[i];
  // do your stuff
  console.log(val);
}
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  2. 0

    If you really do not care about the array. Sort the array and remove them as you use them. There is no need to keep looking up the min.

    function getLeast(arr) {
  const remaining = [...arr].sort()
  return () => {
    if (!remaining.length) return undefined;
    return remaining.shift();
  }
}

const nums = [3, 8, 2, 5, 1];

const genNums = getLeast(nums);
console.log(genNums()); // 1
console.log(genNums()); // 2
console.log(genNums()); // 3
console.log(genNums()); // 5
console.log(genNums()); // 8
console.log(genNums()); // undefined

    using a generator

    function* getLeast(arr) {
  const remaining = [...arr].sort()
  while (remaining.length) {
    yield remaining.shift();
  }
}

const nums = [3, 8, 2, 5, 1];

const genNums = getLeast(nums);
console.log(genNums.next().value); // 1
console.log(genNums.next().value); // 2
console.log(genNums.next().value); // 3
console.log(genNums.next().value); // 5
console.log(genNums.next().value); // 8
console.log(genNums.next().value); // undefined
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