import java.util.*; public class Main { public...

To minimize the number of even‐sized runs (and secondarily the number of bit‐flips you need), you can treat your string as a sequence of pairs and do a tiny DP over those pairs. Here’s the idea:

1. Break the length-n string into $m = n/2$ adjacent pairs $(s[0],s[1]),(s[2],s[3]),\dots$.
2. For each pair $j$ and each target bit $b\in\{0,1\}$, compute

$$\text{cost}[j][b] = (s[2j] \ne b) + (s[2j+1] \ne b),$$

the number of flips to turn both bits of pair $j$ into $b$.
3. Let

$$\text{dpSeg}[j][b],\;\text{dpFlip}[j][b]$$

be the best (min segments, then min flips) you can do on the first $j+1$ pairs if pair $j$ is set to bit $b$.
4. Recurrence for $j>0$:

text
if b == b':  // continuing the same run
  seg  = dpSeg[j-1][b']
  flip = dpFlip[j-1][b'] + cost[j][b]
else:         // starting a new run
  seg  = dpSeg[j-1][b'] + 1
  flip = dpFlip[j-1][b'] + cost[j][b]

Take the $(\text{seg},\text{flip})$ pair that’s lexicographically smallest over b′∈{0,1}.
5. Initialize at $j=0$ with one segment:
$\text{dpSeg}[0][b]=1$, $\text{dpFlip}[0][b]=\text{cost}[0][b]$.
6. At the end pick the better of $\bigl(\text{dpSeg}[m-1][0],\text{dpFlip}[m-1][0]\bigr)$ vs.\ $\bigl(\text{dpSeg}[m-1][1],\text{dpFlip}[m-1][1]\bigr)$, then backtrack to recover where the segment‐breaks were. Each segment you get spans a certain number of pairs; you multiply by 2 to get your even run‐lengths.

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Java Implementation

java
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        int n = s.length();
        if (n % 2 != 0) {
            System.out.println("String length must be even");
            return;
        }
        int m = n / 2;

        // 1) precompute flip‐cost for each pair j into bit 0 or 1
        int[][] cost = new int[m][2];
        for (int j = 0; j < m; j++) {
            char c1 = s.charAt(2*j), c2 = s.charAt(2*j+1);
            cost[j][0] = (c1!='0' ? 1:0) + (c2!='0' ? 1:0);
            cost[j][1] = (c1!='1' ? 1:0) + (c2!='1' ? 1:0);
        }

        // 2) dpSeg[j][b] = min # of segments up to pair j ending in bit b
        //    dpFlip[j][b] = min flips (tie‐broken after dpSeg) for that choice
        int[][] dpSeg  = new int[m][2];
        int[][] dpFlip = new int[m][2];
        // to backtrack: what bit we came from
        int[][] parent = new int[m][2];

        // initialize j=0
        for (int b = 0; b < 2; b++) {
            dpSeg[0][b]  = 1;
            dpFlip[0][b] = cost[0][b];
            parent[0][b] = -1;
        }

        // fill DP
        for (int j = 1; j < m; j++) {
            for (int b = 0; b < 2; b++) {
                int bestSeg = Integer.MAX_VALUE;
                int bestFlip = Integer.MAX_VALUE;
                int bestPb  = -1;
                // try previous bit pb
                for (int pb = 0; pb < 2; pb++) {
                    int seg  = dpSeg[j-1][pb] + (b==pb ? 0 : 1);
                    int flip = dpFlip[j-1][pb] + cost[j][b];
                    // lex compare (seg,flip)
                    if (seg < bestSeg || (seg==bestSeg && flip < bestFlip)) {
                        bestSeg  = seg;
                        bestFlip = flip;
                        bestPb   = pb;
                    }
                }
                dpSeg[j][b]  = bestSeg;
                dpFlip[j][b] = bestFlip;
                parent[j][b] = bestPb;
            }
        }

        // pick the better ending bit
        int endBit = (dpSeg[m-1][0] < dpSeg[m-1][1]
                   || (dpSeg[m-1][0]==dpSeg[m-1][1]
                       && dpFlip[m-1][0] <= dpFlip[m-1][1]))
                     ? 0 : 1;

        // backtrack to find segment boundaries
        List<Integer> segmentsInPairs = new ArrayList<>();
        for (int j = m-1, b = endBit; j >= 0; j--) {
            int pb = parent[j][b];
            // if new segment here?
            if (j == 0 || b != pb) {
                segmentsInPairs.add(1);
            } else {
                segmentsInPairs.set(segmentsInPairs.size()-1,
                                    segmentsInPairs.get(segmentsInPairs.size()-1) + 1);
            }
            b = pb;
        }
        // reverse to get front‐to‐back
        Collections.reverse(segmentsInPairs);

        // each segment is that many pairs → run length = pairs*2
        List<Integer> a = new ArrayList<>();
        for (int pcount : segmentsInPairs) {
            a.add(pcount * 2);
        }

        // output: the run‐length array and the flips used
        System.out.println(a + "  flips=" + dpFlip[m-1][endBit]);
    }
}

How it works

• We only ever create runs whose lengths are multiples of 2.
• The DP minimizes first the total number of runs (segments), and then among all ways to achieve that it minimizes the total number of bit‐flips.
• After backtracking you get a list of how many pairs each run contains; multiplying by 2 gives you your final run‐lengths.

This runs in $O(n)$ time and $O(n)$ space and gives you the smallest possible number of even‐length groups.