Making two sequences increasing in JavaScript

In JavaScript, making two sequences strictly increasing by swapping elements at the same indices is a dynamic programming problem. A sequence is strictly increasing if each element is greater than the previous one.

Problem Statement

Given two arrays arr1 and arr2, we can swap elements at any index i between the arrays. The goal is to find the minimum number of swaps needed to make both arrays strictly increasing.

Understanding the Example

Consider the input arrays:

const arr1 = [1, 3, 5, 4];
const arr2 = [1, 2, 3, 7];

console.log("Original arrays:");
console.log("arr1:", arr1);
console.log("arr2:", arr2);

Original arrays:
arr1: [ 1, 3, 5, 4 ]
arr2: [ 1, 2, 3, 7 ]

Array arr1 is not strictly increasing because 5 > 4. By swapping elements at index 3, we get [1, 3, 5, 7] and [1, 2, 3, 4], both strictly increasing.

Dynamic Programming Solution

We use dynamic programming to track the minimum swaps needed. For each position, we maintain two states: swapped or not swapped at the current index.

const findSwaps = (arr1 = [], arr2 = []) => {
    // Map to store minimum swaps for current position
    // true: swapped at current index, false: not swapped
    let map = {
        true: 1,   // Start with 1 swap if we swap at index 0
        false: 0,  // Start with 0 swaps if we don't swap at index 0
    };
    
    for (let i = 1; i < arr1.length; i++) {
        const current = {
            true: Infinity,
            false: Infinity,
        };
        
        // Check if we can swap at current position
        if (arr1[i] > arr2[i - 1] && arr2[i] > arr1[i - 1]) {
            // We can swap: arr1[i] goes to arr2, arr2[i] goes to arr1
            current.true = Math.min(current.true, map.false + 1);
            current.false = Math.min(current.false, map.true);
        }
        
        // Check if we can keep elements at current position
        if (arr2[i] > arr2[i - 1] && arr1[i] > arr1[i - 1]) {
            // We can keep: both arrays remain increasing
            current.true = Math.min(current.true, map.true + 1);
            current.false = Math.min(current.false, map.false);
        }
        
        map = current;
    }
    
    return Math.min(map.false, map.true);
};

const arr1 = [1, 3, 5, 4];
const arr2 = [1, 2, 3, 7];

console.log("Minimum swaps needed:", findSwaps(arr1, arr2));

Minimum swaps needed: 1

How the Algorithm Works

The algorithm considers two scenarios at each position:

1. Cross-swap scenario: Check if swapping creates valid increasing sequences
2. No-swap scenario: Check if keeping elements maintains increasing sequences

// Test with another example
const testArr1 = [0, 4, 4, 5, 9];
const testArr2 = [0, 1, 6, 8, 10];

console.log("Test arrays:");
console.log("arr1:", testArr1);
console.log("arr2:", testArr2);
console.log("Minimum swaps:", findSwaps(testArr1, testArr2));

Test arrays:
arr1: [ 0, 4, 4, 5, 9 ]
arr2: [ 0, 1, 6, 8, 10 ]
Minimum swaps: 1

Key Points

• Dynamic programming tracks minimum swaps for each position
• Two states are maintained: swapped or not swapped at current index
• Algorithm checks both cross-swap and no-swap possibilities
• Time complexity: O(n), Space complexity: O(1)

Conclusion

This dynamic programming approach efficiently finds the minimum swaps needed to make both sequences strictly increasing. The algorithm considers all valid swap combinations and returns the optimal solution.