Palindrome

index.js

const isPalindrome = str => [...str].reverse().join('') === str

const canBePalindrome = str => {
    const res = [...str].reduce((a,c) => {
        a[c] = !a[c] ? 1 : a[c]+1
        return a
    }, {})
    //there needs to be 1 odd number letter and the reset even
    const check = Object.entries(res).map(x => {
        return x[1] % 2
    }).reduce((a,c) => a+c)
    // Check should contain the number of letters that are odd numbered
    // if > 1 then it's a no go
    return check === 1
}
// is pal
console.log('isPalindrome()')
console.log(`\tracecar should be true`, isPalindrome(`racecar`))
console.log(`\tracec should be false`, isPalindrome(`racec`))

// can be pal
console.log('canBePalindrome()')
console.log(`\tracecar should be true`, canBePalindrome(`racecar`))
console.log(`\trraacce should be true`, canBePalindrome(`rraacce`))
console.log(`\tracec should be false`, canBePalindrome(`racec`))
console.log(`\taabbcbbaa should be true`, canBePalindrome(`aabbcbbaa`))
console.log(`\taabbc should be true`, canBePalindrome(`aabbc`))
console.log(`\tabc should be false`, canBePalindrome(`racec`))

Results

isPalindrome()
	racecar should be true true
	racec should be false false
canBePalindrome()
	racecar should be true true
	rraacce should be true true
	racec should be false false
	aabbcbbaa should be true true
	aabbc should be true true
	abc should be false false