LeetCode #1505 — HARD

Minimum Possible Integer After at Most K Adjacent Swaps On Digits

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given a string num representing the digits of a very large integer and an integer k. You are allowed to swap any two adjacent digits of the integer at most k times.

Return the minimum integer you can obtain also as a string.

Example 1:

Input: num = "4321", k = 4
Output: "1342"
Explanation: The steps to obtain the minimum integer from 4321 with 4 adjacent swaps are shown.

Example 2:

Input: num = "100", k = 1
Output: "010"
Explanation: It's ok for the output to have leading zeros, but the input is guaranteed not to have any leading zeros.

Example 3:

Input: num = "36789", k = 1000
Output: "36789"
Explanation: We can keep the number without any swaps.

Constraints:

1 <= num.length <= 3 * 10⁴
num consists of only digits and does not contain leading zeros.
1 <= k <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given a string num representing the digits of a very large integer and an integer k. You are allowed to swap any two adjacent digits of the integer at most k times. Return the minimum integer you can obtain also as a string.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Greedy · Segment Tree

Example 1

"4321"
4

Example 2

"100"
1

Example 3

"36789"
1000

Step 02

Core Insight

What unlocks the optimal approach

We want to make the smaller digits the most significant digits in the number.
For each index i, check the smallest digit in a window of size k and append it to the answer. Update the indices of all digits in this range accordingly.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1505: Minimum Possible Integer After at Most K Adjacent Swaps On Digits
class Solution {
    public String minInteger(String num, int k) {
        Queue<Integer>[] pos = new Queue[10];
        for (int i = 0; i < 10; ++i) {
            pos[i] = new ArrayDeque<>();
        }
        int n = num.length();
        for (int i = 0; i < n; ++i) {
            pos[num.charAt(i) - '0'].offer(i + 1);
        }
        StringBuilder ans = new StringBuilder();
        BinaryIndexedTree tree = new BinaryIndexedTree(n);
        for (int i = 1; i <= n; ++i) {
            for (int v = 0; v < 10; ++v) {
                if (!pos[v].isEmpty()) {
                    Queue<Integer> q = pos[v];
                    int j = q.peek();
                    int dist = tree.query(n) - tree.query(j) + j - i;
                    if (dist <= k) {
                        k -= dist;
                        q.poll();
                        ans.append(v);
                        tree.update(j, 1);
                        break;
                    }
                }
            }
        }
        return ans.toString();
    }
}

class BinaryIndexedTree {
    private int n;
    private int[] c;

    public BinaryIndexedTree(int n) {
        this.n = n;
        c = new int[n + 1];
    }

    public void update(int x, int delta) {
        while (x <= n) {
            c[x] += delta;
            x += lowbit(x);
        }
    }

    public int query(int x) {
        int s = 0;
        while (x > 0) {
            s += c[x];
            x -= lowbit(x);
        }
        return s;
    }

    public static int lowbit(int x) {
        return x & -x;
    }
}

// Accepted solution for LeetCode #1505: Minimum Possible Integer After at Most K Adjacent Swaps On Digits
type BinaryIndexedTree struct {
	n int
	c []int
}

func newBinaryIndexedTree(n int) *BinaryIndexedTree {
	c := make([]int, n+1)
	return &BinaryIndexedTree{n, c}
}

func (this *BinaryIndexedTree) lowbit(x int) int {
	return x & -x
}

func (this *BinaryIndexedTree) update(x, delta int) {
	for x <= this.n {
		this.c[x] += delta
		x += this.lowbit(x)
	}
}

func (this *BinaryIndexedTree) query(x int) int {
	s := 0
	for x > 0 {
		s += this.c[x]
		x -= this.lowbit(x)
	}
	return s
}

func minInteger(num string, k int) string {
	pos := make([][]int, 10)
	for i, c := range num {
		pos[c-'0'] = append(pos[c-'0'], i+1)
	}
	n := len(num)
	tree := newBinaryIndexedTree(n)
	var ans strings.Builder
	for i := 1; i <= n; i++ {
		for v := 0; v < 10; v++ {
			if len(pos[v]) > 0 {
				j := pos[v][0]
				dist := tree.query(n) - tree.query(j) + j - i
				if dist <= k {
					k -= dist
					pos[v] = pos[v][1:]
					ans.WriteByte(byte(v + '0'))
					tree.update(j, 1)
					break
				}
			}
		}
	}
	return ans.String()
}

# Accepted solution for LeetCode #1505: Minimum Possible Integer After at Most K Adjacent Swaps On Digits
class BinaryIndexedTree:
    def __init__(self, n):
        self.n = n
        self.c = [0] * (n + 1)

    @staticmethod
    def lowbit(x):
        return x & -x

    def update(self, x, delta):
        while x <= self.n:
            self.c[x] += delta
            x += BinaryIndexedTree.lowbit(x)

    def query(self, x):
        s = 0
        while x:
            s += self.c[x]
            x -= BinaryIndexedTree.lowbit(x)
        return s


class Solution:
    def minInteger(self, num: str, k: int) -> str:
        pos = defaultdict(deque)
        for i, v in enumerate(num, 1):
            pos[int(v)].append(i)
        ans = []
        n = len(num)
        tree = BinaryIndexedTree(n)
        for i in range(1, n + 1):
            for v in range(10):
                q = pos[v]
                if q:
                    j = q[0]
                    dist = tree.query(n) - tree.query(j) + j - i
                    if dist <= k:
                        k -= dist
                        q.popleft()
                        ans.append(str(v))
                        tree.update(j, 1)
                        break
        return ''.join(ans)

// Accepted solution for LeetCode #1505: Minimum Possible Integer After at Most K Adjacent Swaps On Digits
/**
 * [1505] Minimum Possible Integer After at Most K Adjacent Swaps On Digits
 *
 * You are given a string num representing the digits of a very large integer and an integer k. You are allowed to swap any two adjacent digits of the integer at most k times.
 * Return the minimum integer you can obtain also as a string.
 *  
 * Example 1:
 * <img alt="" src="https://assets.leetcode.com/uploads/2020/06/17/q4_1.jpg" style="width: 500px; height: 40px;" />
 * Input: num = "4321", k = 4
 * Output: "1342"
 * Explanation: The steps to obtain the minimum integer from 4321 with 4 adjacent swaps are shown.
 *
 * Example 2:
 *
 * Input: num = "100", k = 1
 * Output: "010"
 * Explanation: It's ok for the output to have leading zeros, but the input is guaranteed not to have any leading zeros.
 *
 * Example 3:
 *
 * Input: num = "36789", k = 1000
 * Output: "36789"
 * Explanation: We can keep the number without any swaps.
 *
 *  
 * Constraints:
 *
 * 	1 <= num.length <= 3 * 10^4
 * 	num consists of only digits and does not contain leading zeros.
 * 	1 <= k <= 10^9
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/minimum-possible-integer-after-at-most-k-adjacent-swaps-on-digits/
// discuss: https://leetcode.com/problems/minimum-possible-integer-after-at-most-k-adjacent-swaps-on-digits/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    // Credit: https://leetcode.com/problems/minimum-possible-integer-after-at-most-k-adjacent-swaps-on-digits/solutions/3162681/just-a-runnable-solution/
    pub fn min_integer(num: String, k: i32) -> String {
        let mut k = k as usize;
        let num = num.into_bytes();
        let n = num.len();
        let mut result = Vec::with_capacity(n);
        let mut q = vec![n; 10];
        for (i, &item) in num.iter().enumerate() {
            let d = (item - b'0') as usize;
            if q[d] == n {
                q[d] = i;
            }
        }
        let mut used = vec![false; n];
        let mut q_used = vec![0; 10];

        for _ in 0..n {
            for d in 0..10_usize {
                if q[d] == n {
                    continue;
                }
                let c = q[d] - q_used[d];
                if c <= k {
                    k -= c;
                    result.push(b'0' + d as u8);
                    used[q[d]] = true;
                    for d1 in 0..10_usize {
                        if q[d1] > q[d] {
                            q_used[d1] += 1;
                        }
                    }
                    while q[d] < n {
                        if used[q[d]] {
                            q_used[d] += 1;
                        }
                        q[d] += 1;
                        let &c = num.get(q[d]).unwrap_or(&0_u8);
                        if c == b'0' + d as u8 {
                            break;
                        }
                    }
                    break;
                }
            }
        }

        String::from_utf8(result).unwrap()
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_1505_example_1() {
        let num = "4321".to_string();
        let k = 4;

        let result = "1342".to_string();

        assert_eq!(Solution::min_integer(num, k), result);
    }

    #[test]
    fn test_1505_example_2() {
        let num = "36789".to_string();
        let k = 1000;

        let result = "36789".to_string();

        assert_eq!(Solution::min_integer(num, k), result);
    }

    #[test]
    fn test_1505_example_3() {
        let num = "100".to_string();
        let k = 1;

        let result = "010".to_string();

        assert_eq!(Solution::min_integer(num, k), result);
    }
}

// Accepted solution for LeetCode #1505: Minimum Possible Integer After at Most K Adjacent Swaps On Digits
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1505: Minimum Possible Integer After at Most K Adjacent Swaps On Digits
// class Solution {
//     public String minInteger(String num, int k) {
//         Queue<Integer>[] pos = new Queue[10];
//         for (int i = 0; i < 10; ++i) {
//             pos[i] = new ArrayDeque<>();
//         }
//         int n = num.length();
//         for (int i = 0; i < n; ++i) {
//             pos[num.charAt(i) - '0'].offer(i + 1);
//         }
//         StringBuilder ans = new StringBuilder();
//         BinaryIndexedTree tree = new BinaryIndexedTree(n);
//         for (int i = 1; i <= n; ++i) {
//             for (int v = 0; v < 10; ++v) {
//                 if (!pos[v].isEmpty()) {
//                     Queue<Integer> q = pos[v];
//                     int j = q.peek();
//                     int dist = tree.query(n) - tree.query(j) + j - i;
//                     if (dist <= k) {
//                         k -= dist;
//                         q.poll();
//                         ans.append(v);
//                         tree.update(j, 1);
//                         break;
//                     }
//                 }
//             }
//         }
//         return ans.toString();
//     }
// }
// 
// class BinaryIndexedTree {
//     private int n;
//     private int[] c;
// 
//     public BinaryIndexedTree(int n) {
//         this.n = n;
//         c = new int[n + 1];
//     }
// 
//     public void update(int x, int delta) {
//         while (x <= n) {
//             c[x] += delta;
//             x += lowbit(x);
//         }
//     }
// 
//     public int query(int x) {
//         int s = 0;
//         while (x > 0) {
//             s += c[x];
//             x -= lowbit(x);
//         }
//         return s;
//     }
// 
//     public static int lowbit(int x) {
//         return x & -x;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.