Method 1: Using a New Array for Comparison

This approach iterates through the original array and checks each element against a new array. If the element is not found, it is added.

const originalArray = ['x', 5, 5, 5, 7, 9, 9, 'y', 'z', 'x'];
const uniqueArray = [];
const length = originalArray.length;

for (let i = 0; i < length; i++) {
    let isDuplicate = false;
    for (const item of uniqueArray) {
        if (item === originalArray[i]) {
            isDuplicate = true;
            break;
        }
    }
    if (!isDuplicate) {
        uniqueArray.push(originalArray[i]);
    }
}
console.log('Result from Method 1:', uniqueArray);

Method 2: In-Place Deletion with Nested Loops

This method modifies the original array by comparing elements and removing duplciates using splice.

let dataSet = [10, 20, 30, 'p', 'q', 30, 20, 'r', 'p'];
let currentLength = dataSet.length;

while (currentLength-- > 0) {
    for (let j = 0; j < currentLength; j++) {
        if (dataSet[j] === dataSet[currentLength]) {
            dataSet.splice(j, 1);
            break;
        }
    }
}
console.log('Result from Method 2:', dataSet);

Method 3: Utilziing indexOf for Position Comparison

Elements are removed in-place if their first occurrence index does not match the current iteration index.

const numberList = [15, 42, 15, 15, 15, 8, 42, 12, 18, 21, 21, 17, 12];
function eliminateDuplicates(arr) {
    for (let idx = 0; idx < arr.length; idx++) {
        if (arr.indexOf(arr[idx]) !== idx) {
            arr.splice(idx, 1);
            idx--;
        }
    }
    return arr;
}
const processedList = eliminateDuplicates(numberList);
console.log('Result from Method 3:', processedList);

Method 4: Building a New Array with indexOf

A new array is constructed by checking if elements exist within it using indexOf.

const mixedValues = ['g', 'h', 100, 200, 300, 300, 300, 'k', 'g'];
const filteredValues = [];

for (let k = 0; k < mixedValues.length; k++) {
    if (filteredValues.indexOf(mixedValues[k]) === -1) {
        filteredValues.push(mixedValues[k]);
    }
}
console.log('Result from Method 4:', filteredValues);

Method 5: Leveraging the filter Method

A concise solution using filter that retains only the first occurrence of each element.

const sampleArray = [22, 33, 44, 22, 33, 55, 66];
const deduplicatedArray = sampleArray.filter((value, position, self) => {
    return self.indexOf(value) === position;
});
console.log('Result from Method 5:', deduplicatedArray);