Removing least number of elements to convert array into increasing sequence using JavaScript

We need to write a JavaScript function that removes the minimum number of elements from an array to make it strictly increasing. This is essentially finding the Longest Increasing Subsequence (LIS) and removing all other elements.

Problem Analysis

The goal is to find the longest possible increasing subsequence and remove the fewest elements. A strictly increasing sequence means each element is greater than the previous one.

Example with Corrected Algorithm

The original approach has flaws - it doesn't find the optimal solution. Here's a correct implementation using dynamic programming:

const arr = [1, 100, 2, 3, 100, 4, 5];

const findIncreasingArray = (arr = []) => {
    if (arr.length <= 1) return arr;
    
    // Find LIS using dynamic programming
    const lis = new Array(arr.length).fill(1);
    const parent = new Array(arr.length).fill(-1);
    
    for (let i = 1; i < arr.length; i++) {
        for (let j = 0; j < i; j++) {
            if (arr[j] < arr[i] && lis[j] + 1 > lis[i]) {
                lis[i] = lis[j] + 1;
                parent[i] = j;
            }
        }
    }
    
    // Find the index with maximum LIS length
    let maxLength = Math.max(...lis);
    let maxIndex = lis.indexOf(maxLength);
    
    // Reconstruct the LIS
    const result = [];
    let current = maxIndex;
    while (current !== -1) {
        result.unshift(arr[current]);
        current = parent[current];
    }
    
    return result;
};

console.log("Original array:", arr);
console.log("Increasing sequence:", findIncreasingArray(arr));
console.log("Elements removed:", arr.length - findIncreasingArray(arr).length);

Original array: [1, 100, 2, 3, 100, 4, 5]
Increasing sequence: [1, 2, 3, 4, 5]
Elements removed: 2

Simplified Greedy Approach

For a simpler but less optimal solution, we can use a greedy approach that builds the sequence left-to-right:

const findIncreasingArrayGreedy = (arr = []) => {
    if (arr.length === 0) return [];
    
    const result = [arr[0]];
    
    for (let i = 1; i < arr.length; i++) {
        if (arr[i] > result[result.length - 1]) {
            result.push(arr[i]);
        }
    }
    
    return result;
};

const testArray = [1, 100, 2, 3, 100, 4, 5];
console.log("Greedy result:", findIncreasingArrayGreedy(testArray));

Greedy result: [1, 100]

Comparison

Key Points

• The dynamic programming approach guarantees the minimum number of removals
• The greedy approach is faster but may not find the optimal solution
• This problem is equivalent to finding the Longest Increasing Subsequence (LIS)

Conclusion

Use dynamic programming for the optimal solution that removes the fewest elements. The greedy approach works faster but may remove more elements than necessary.