Contains Duplicate - JS Solution

Editorial

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const CURRENT = 'current', CURRENT_COLOR = 'orange';
const FOUND = 'found', FOUND_COLOR = 'green';

function visMainFunction(inputNums) {
    initClassProperties();

    startBatch();
    const nums = new VisArray(...inputNums);
    const seenNums = new VisSet();
    let index = 0;
    makeVisVariable(index).registerAsIndexPointer(nums, CURRENT);
    endBatch();

    explanationAndColorPaletteNotification();

    for (let num of nums) {
        startPartialBatch();

        if (seenNums.has(num)) {
            seenNums.makeVisManagerForMember(num).addClass(FOUND);
            notification(`
              ✅ ${makeColoredSpan(num, FOUND_COLOR)} is found in <b>seenNums</b> set, meaning it was seen before.<br/>
              Therefore, return <b>true</b>.
            `);
            return true;
        }

        notification(`
          ❌ ${makeColoredSpan(num, CURRENT_COLOR)} is missing in <b>seenNums</b> set, meaning it has not been seen before.<br/>
          ➕ Add it to <b>seenNums</b> to mark it as seen. If it will be encountered again, we will return <b>true</b> immediately.
        `);
        seenNums.add(num);

        index += 1;
        endBatch();
    }

    notification(`
      📦 No duplicate is found in the array.<br/>
      Return <b>false</b>.
    `);
    return false;
}

function explanationAndColorPaletteNotification() {
    explanationNotification();
    notification(`
        🎨 <b>Color Palette Guide</b><br/><br/>
        • ${makeColoredSpan("Orange", CURRENT_COLOR)} background → <b>Current number</b><br/>
        • ${makeColoredSpan("Green", FOUND_COLOR)} background → <b>Duplicate found</b><br/>
    `);
}

function initClassProperties() {
   setClassProperties(CURRENT, {
    'backgroundColor': CURRENT_COLOR
   });

   setClassProperties(FOUND, {
    'backgroundColor': FOUND_COLOR
   });
}

Approach

Goal

Given an array of integers, determine whether any value appears at least twice.

Key Idea (Intuition)

Checking against the previously seen numbers is enough: if the current number has been seen before, we have found a duplicate. Passing a number once does not mean it cannot be a duplicate — we may still encounter it later, and when that happens, the new number is checked against all those seen earlier and the duplicate will be detected. To enable this, we maintain a set seenNums of numbers encountered so far and check in constant time whether the current number has already been seen.

Traversal

Iterate through the array from left to right, examining one number at a time.

Per-Iteration Logic

Duplicate check: If the current number is already in seenNums, return true immediately.
Record as seen: Otherwise, add the current number to seenNums and continue.

Return

If the loop finishes without finding a repeated number, return false.

Time Complexity

The array is traversed once from left to right. At each iteration, checking whether a number is in seenNums and adding a number to seenNums both take amortized constant time. Therefore, the overall time complexity is O(n), where n is the number of elements in the array.

Space Complexity

The algorithm uses a set seenNums to store previously encountered numbers. In the worst case, if all numbers are distinct, the set grows to the size of the array. Therefore, the space complexity is O(n).

Input-1

Contains Duplicate - JS Solution

Solution code

Solution explanation

Solution explanation

Approach

Goal

Key Idea (Intuition)

Traversal

Per-Iteration Logic

Return

Time Complexity

Space Complexity